THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


"i|C  r  'ast  date  stamped  below 


SOUTHERN    BRANCH 

UNIVERSITY  OF  CALIFORNIA 
LIBRARY 

LOS  ANGELES.  CALIF, 


TREATISE 


ON 


PLANE  AND  SPHERICAL 


TRIGONOMETRY. 


BY 
WILLIAM    CHAUVENET, 

PROFESSOR  OF  MATHEMATICS  AND  ASTRONOMY   IN  WASHINGTON  UNIVERSITY, 
SAINT  LOUIS. 


ELEVENTH   EDITION. 


PHILADELPHIA : 

J.  B.  LIPPINCOTT   COMPANY. 
3 


Entered  according  to  Act  of  Congress,  in  the  year  1850,  by 

WILLIAM   CHAUVENET, 
in  the  Clerk's  Office  ol  the  District  Court  of  the  Eastern  District  of  Pennsjivamc 


ELECTROTYPEO    AND    PRINTED    BY   J.   B.   LIPPINCOTT    COMPANY,   PHILADELPHIA,  U.  8.  A. 


C3<J 
PREFACE. 


I  HAVE  in  this  treatise  endeavored  to  arrange  a  course  of  trigo- 
nometrical study  sufficiently  extensive  to  enable  the  student  to  com- 
prehend readily  any  applications  of  trigonometry  he  may  meet  with 
in  the  works  of  the  best  modern  mathematicians.  With  this  object, 
some  topics  have  been  introduced  which  are  not  usually  found  in 
works  devoted  specially  to  this  subject. 

Among:  those  topics,  the  most  important  is  the  solution  of  the 
general  spherical  triangle,  or  the  triangle  whose  sides  and  angles  are 
not  limited,  according  to  the  usual  practice,  to  values  less  than  180°. 
The  advantage  of  introducing  such  triangles  into  astronomical  inves- 
tigations is  sufficiently  shown  in  the  applications  made  of  them  in  the 
works  of  BESSEL  and  other  German  mathematicians;  and  especially 
in  the  Theoria  Motus  Corporum  Codestium  of  GAUSS,  who  was  the 
first  to  suggest  their  employment. 

The  subject  of  Finite  Differences  of  triangles,  plane  and  spherical, 
occupies  a  large  space  in  Cagnoli's  treatise,  but  has  not  been  admitted 
into  more  recent  works.  It  here  occupies  only  a  few  pages,  but  no 
important  result  of  Cagnoli's  Table  has  been  omitted,  while  a  number 
of  the  formulae  are  much  simpler  than  the  corresponding  ones  given 
by  him. 

Although  my  plan  embraces  a  much  more  extensive  course  than  is 
contained  in  the  text-books  commonly  used,  I  have  studiously  kept 
in  view  the  wants  of  academic  and  collegiate  classes ;  and  have  so 
arranged  the  work  that  a  selection  of  subjects  of  immediate  impor- 
tance may  be  readily  made.  The  more  elementary  portions  are 


4  PREFACE. 

printed  in  a  larger  type,  and  are  intended  to  form,  independently  of 
the  matter  in  the  smaller  type,  a  connected  treatise  which  may  be 
studied  as  though  it  were  in  a  separate  volume. 

Those  who  may  afterwards  wish  to  extend  their  knowledge  will 
appreciate  the  advantage  of  having  the  higher  departments  of  the 
subject  treated  in  connection  with  those  fundamental  ones  to  which 
they  are  most  intimately  related. 

w.  c. 

U.  S.  NAVAL  ACADEMY, 

Annapolis,  Md.,  May  1,  1850. 


NOTE  TO  THE  FOURTH  EDITION. 


In  this  edition,  besides  a  number  of  minor  changes,  and  the  correc- 
tion of  some  typographical  errors,  a  very  important  modification  has 
been  made  in  the  solution  of  the  equation  tan  x=p  tan  y  by  series 
(p.  145),  which  was  given  in  former  editions  in  the  usual  form  as 
stated  by  all  writers  on  trigonometry.  This  form  was  discovered  to 
lack  generality,  and  consequently  to  fail  in  certain  applications,  in 
consequence  of  the  omission  of  the  arbitrary  term  m:  now  introduced. 
Several  subsequent  investigations,  depending  on  this,  have  in  like 

manner  been  rectified. 

W.  C. 

U.  S.  NAVAL  ACADEMY,  April  1,  1854. 


CONTENTS. 


PART  I. 
PLANE  TRIGONOMETRY. 

CHAPTER  I. 

FAGS 

MEASURES  OF  ANGLES  AND  ARCS 9 

CHAPTER  II. 
SINES,  TANGENTS,  AND  SECANTS.    FUNDAMENTAL  FORMULAS _14 

CHAPTER   III. 

TRIGONOMETRIC  FUNCTIONS  OF  ANGULAR  MAGNITUDE  IN  GENERAL  ....     22 
Sine  and  Tangent  of  a  Small  Angle  or  Arc 30 

CHAPTER  IV. 

GENERAL  FORMULAE 31 

Formulae  for  Multiple  Angles JJ8 

Relations  of  Three  Angles 38 

Inverse  Trigonometric  Functions 41 

CHAPTER  V. 

TRIGONOMETRIC  TABLES 43 

Elementary  Method  of  Constructing  the  Trigonometric  Table 47 

CHAPTER  VI. 

SOLUTION  OF  PLANE  RIGHT  TRIANGLES 51 

Additional  Formula;  for  Right  Triangles 54 

CHAPTER  VII. 

FORMULAE  FOR  THE  SOLUTION  OF  PLANE  OBLIQUE  TRIANGLES 67 

A2  6 


6  CONTENTS. 

CHAPTER  VIII. 

PAGE 

SOLUTION  OF  PLANE  OBLIQUE  TRIANGLES 64 

Area  of  a  Plane  Triangle 74 

CHAPTEE  IX. 
MISCELLANEOUS  PROBLEMS  RELATING  TO  PLANE  TRIANGLES 75 

CHAPTER  X. 

SOLUTION  OF  CERTAIN  TRIGONOMETRIC  EQUATIONS  AND  OF  NUMERICAL  EQUA- 
TIONS OF  THE  SECOND  AND  THIRD  DEGREES 85 

CHAPTER  XI. 

DIFFERENCES  AND  DIFFERENTIALS  OF  THE  TRIGONOMETRIC  FUNCTIONS   .    .    101 

CHAPTER  XII. 
DIFFERENCES  AND  DIFFERENTIALS  OF  PLANE  TRIANGLES 105 

CHAPTER  XIII. 

TRIGONOMETRIC  SERIES.     DEVELOPMENTS  OF  THE  FUNCTIONS  OF  AN  ANGLE 

IN  TERMS  OF  THE  ARC,  AND  RECIPROCALLY 115 

Computation  of  Natural  Sines  and  Cosines  by  Series 116 

Computation  of  the  Ratio  of  the  Circumference  of  a  Circle  to  its  Diameter  .    120 
Computation  of  Logarithmic  Sines  and  Cosines 122 

CHAPTER  XIV. 
EXPONENTIAL  FORMULA.    TRINOMIAL  OR  QUADRATIC  FACTORS 127 

CHAPTER   XV. 

TRIGONOMETRIC  SERIES  CONTINUED.  MULTIPLE  ANGLES 135 

Development  of  the  Sine  and  Cosine  of  the  Multiple  Angle  in  a  Series  of 

Ascending  Powers  of  the  Cosine  of  the  Simple  Angle 137 

Development  of  the  Sine  and  Cosine  of  the  Multiple  Angle  in  a  Series  of 

Ascending  Powers  of  the  Sine  of  the  Simple  Angle 139 

Development  of  the  Sine  and  Cosine  of  the  Multiple  Angle  in  a  Series  of 

Ascending  Powers  of  the  Tangent  of  the  Simple  Angle 140 

Development  of  any  power  of  the  Cosine  of  the  Simple  Angle  in  a  Series 

of  Sines  or  Cosines  of  the   Multiple  Angles,  the  Cosine  of  the   Simple 

Angle  being  positive 141 

Development  of  any  power  of  the  Cosine  of  the  Simple  Angle  in  a  Series 

of  Sines  or  Cosines  of  the  Multiple  Angles,  the  Cosine  of  the  Simple  Angle 

being  negative 142 

Development  of  any  power  of  the  Sine  of  the  Simple  Angle  in  a  Series  of 

Sines  or  Cosines  of  the  Multiple  Angles 144 

Certain  Equations  developed  in  Series  of  Multiple  Angles 145 


CONTENTS.  7 

PART  II. 
SPHERICAL  TRIGONOMETRY. 

CHAPTER  I. 

PAGE 

GENERAL  FORMULA 149 

Gauss's  Theorem 161 

Additional  Formula? 162 

Deduction  of  the  Formulae   of  Plane   Triangles   from   those  of  Spherical 

Triangles 166 

CHAPTER  II. 

SOLUTION  OF  SPHERICAL  RIGHT  TRIANGLES 167 

Additional  Formulae  for  the  Solution  of  Spherical  Right  Triangles    .    .    .    176 
Quadrantal  and  Isosceles  Triangles 177 

CHAPTER  III. 

SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES 178 

Solution  of  Spherical  Oblique  Triangles  by  means  of  a  Perpendicular   .    .    206 
Computation  of  Spherical  Formulae  by  the  Gaussian  Table 211 

CHAPTER  IV. 

SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE 214 

Note  upon  Gauss's  Equations 227 

CHAPTER  V. 
AREA  OF  A  SPHERICAL  TRIANGLE 229 

CHAPTER  VI. 
DIFFERENCES  AND  DIFFERENTIALS  OF  SPHERICAL  TRIANGLES 232 

CHAPTER  VII. 

APPROXIMATE  SOLUTION  OF  SPHERICAL  TRIANGLES  IN  CERTAIN  CASES   .   .    241 
Legendre's  Theorem 244 

CHAPTER  VIII. 
MISCELLANEOUS  PROBLEMS  OF  SPHERICAL  TRIGONOMETRY    .  ...    246 


PART  I. 
PLANE  TRIGONOMETRY. 


CHAPTER    I. 
MEASURES  OF  ANGLES  AND  ARCS. 

1.  TRIGONOMETRY  is  that  branch  of  Mathematics  which  treats 
of  methods  of  subjecting  angles  and  triangles  to  numerical  compu- 
tation. 

2.  PLANE  TRIGONOMETRY  treats  of  methods  of  computing  plane 
angles  and  triangles. 

It  embraces  the  investigation  of  the  relations  of  angles  in  gene- 
ral, a  branch  of  the  science  not  necessarily  connected  with  the 
elementary  solution  of  triangles,  and  which  has  been  distinguished 
as  the  Angular  Analysis. 

3.  By  the  solution  of  a  triangle,  in  trigonometry,  is   meant  the 
computation  of  unknown  parts  of  the  triangle  from  given  ones. 

The  triangle  has  six  parts ;  three  angles  and  three  sides.  It  is 
shown  in  geometry,  that  when  any  three  of  these  parts  are  given, 
provided  one  of  them  is  a  side,  the  triangle  may  be  constructed, 
and  the  unknown  parts  found  by  mechanical  measurement. 

In  the  same  cases,  by  trigonometry,  we  compute  the  unknown 
parts  from  the  three  given  ones,  without  resorting  to  construction 
and  measurement :  a  method  of  inferior  accuracy,  on  account  of  the 
unavoidable  imperfections  of  the  instruments  employed,  and  the 
difficulty  of  distinguishing  with  the  eye  the  smallest  subdivisions  of 
lines  and  angles. 

But  here  also  the  case  is  excluded  in  which  the  three  angles  are 
given  without  a  side,  because  there  may  be  an  indefinite  number  of 
plane  triangles,  whose  angles  are  equal  to  the  same  three  given  ones,  as 
2  9 


10 


PLANE  TRIGONOMETRY. 


FIG.  i.  in  Fig.  1,  the  triangles  ABC,  A'B'C', 

etc.  In  this  case,  all  these  triangles 
are  similar,  and  their  sides  are  pro- 
portional ;  or  the  ratio  of  A  B  to  A  C 
is  equal  to  the  ratio  of  A'  B'  to  A'  C', 
etc. ;  so  that  the  ratios  of  the  sides  to 

each  other  are  fixed  or  determinate,  although  the  absolute  lengths 

of  these  sides  are  indeterminate. 

4.  Now,  in  order  to  subject  a  triangle  to  computation,  we  must 
first  express  the  sides  and  angles  by  numbers.      For  this  purpose 
proper  units  of  measure  must  be  adopted. 

The  unit  of  measure  for  the  sides  of  plane  triangles  is  a  straight 
line,  as  an  inch,  a  foot,  a  mile,  etc.;  and  the  number  expressing  a 
side  is  the  number  of  units  of  the  adopted  kind  that  the  side  con- 
tains. 

5.  The  units  by  which  angles  are  expressed  are,  the  degree,  minute, 
and  second;  distinguished  by  the  characters  °  '  ". 

A  degree  is  an  angle  equal  to  -fa  of  a  right  angle  ;  or  a  degree  is 
of  the  whole  angular  space  about  a  point,  or  -^-^  of  four  right 
FIQ.  2.  angles.     Thus,  Fig.  2,  if  the  angular  space  about 

0  is  divided  into  360  equal  parts,  of  which  AO  B 
is  one,  then  AO  B  is  one  degree.  The  right  angle 
will  be  expressed  by  90°  ;  two  right  angles  by 
180°,  and  the  whole  angular  space  about  a  point 
by  360°. 

A*  A  minute  is  an   angle  equal  to  -fa  of  a  degree. 

Therefore,  1°  =  60' ;  and  a  right  angle  =  90  X  60'  =  5400'. 

A  second  is  an  angle  equal  to  -g1^  of  a  minute.  Therefore,  1'  = 
60";  1°  =  60X  60"  =  3600";  and  a  right  angle  =  90  X  60  X  60" 
=  324000". 

Angles  less  than  seconds  are  sometimes  expressed  by  thirds, 
fourths,  fifths,  etc.,  marked  "'  IT  v,  etc.;  a  third  being  fa  of  a  second ; 
a  fourth,  -fa  of  a  third  ;  etc.  But  the  more  convenient  method  is  to 
express  them  as  decimal  parts  of  a  second  ;  thus  \  of  a  right  angle 
will  be  either 

12°  51'  25"  42'"  5PV,  ete. 


_B 
—A 


or  more  conveniently 


12°  51'  25"-714,  ete. 


6.  The  above  division  of  angles  is  called  sexagesimal,  from  the  divisor  60  employed 
in  the  subdivision  of  the  degree.  The  centesimal  division,  however,  would  be  prefer- 
able in  all  cases,  but  cannot  now  be  generally  introduced  without,  at  the  same  time, 
changing  the  arrangement  of  all  our  tables,  the  graduation  of  astronomical  and 


MEASURES  OF  AKCS.  11 

other  instruments,  charts,  etc.  Nevertheless,  the  attempt  has  been  made  in  France, 
and  several  standard  works  exist  in  the  French  language,  in  which  it  is  employed 
throughout. 

Jn  the  centesimal  or  French  division,  the  right  angle  is  divided  into  100  degrees; 
the  degree  into  100  minutes  ;  the  minute  into  100  seconds,  etc.  The  reduction  of 
these  denominations  from  one  to  the  other  requires  only  a  change  in  the  position 
of  the  decimal  point;  thus,  in  this  system  60°  15'  84"  '8  is  the  same  as  607584/A8 
or  60°'75848  or  O9  '6075848,  the  symbol  q  denoting  a  quadrant  or  right  angle. 

To  convert  centesimal  into  sexagesimal  degrees,  since  100°  dec.  =  90°  sex.  deduct  one- 
tenth  from  the  number  of  centesimal  degrees. 

EXAMPLE.     Required  the  number  of  sex.  degrees  in  85°  47'  43"  dec. 

85°'4743  cent. 
Deduct  iV  =  _8  "54743 

7G°'9'2087  sex.  degrees  and  dec.  parts. 


"- 


3G 

or  76°  55'  3G/A732  sexagesimal. 

To  convert  sexagesimal  into  centesimal  degrees,  since  we  must  take  -^  of  the  sex.,  divide 
by  9  and  move  the  decimal  point  one  place  to  the  right. 

EXAMPLE.     Required  the  number  of  centesimal  degrees  in  76°  55X  36"'732  sex. 
Reducing  the  minutes  and  seconds  to  the  decimal  of  a  degree,  we  have 

76°-92687  sex. 

-V-  of  which  is  85°'4743  cent. 

or  85°  47'  43"  centesimal. 

To  distinguish  the  degrees  of  the  centesimal  from  those  of  the  sexagesimal  division, 
the  former  are  frequently  called  grades,  and  are  denoted  by  the  character  *  instead 
of  °  ;  thus  the  preceding  angle  would  be  85e  47'  43//. 


MEASURES  OF  AKCS. 

7.  Since  the  angles  at  the  center  of  a  circle  are  proportional  to  the 
arcs  of  the  circumference  intercepted  between  their  sides,  these  arcs 
may  be  taken  as  the  measures  of  the  angles,  and  we  may  express 
both  the  arc  and  the  angle  by  the  number  of  units  of  arc  intercepted 
on  the  circumference. 

The  units  of  arc  are  also  the  degree,  minute,  and  second.  They 
are  the  arcs  which  subtend  angles  of  a  degree,  a  minute,  and  a  sec- 
ond, respectively,  at  the  center.  A  degree  of  arc  is  thus  always 
^iir  °f  *ne  circumference,  whatever  the  radius  of  the  circle  may  be  : 
and  we  obtain  the  same  numerical  expression  of 
an  angle,  whether  we  refer  it  directly  to  the  angu- 
lar unit,  or  to  the  corresponding  unit  of  arc.  The 
right  angle  AOA',  Fig.  3,  and  its  measure,  the 
quadrant  A  A',  are  therefore  both  expressed  by 
90°  ;  the  semi-circumference  by  180°,  and  the 
whole  circumference  by  360°. 

8.  The  radius  of  the  circle  employed  in  measuring  angles  is  then 


12  PLANE  TRIGONOMETRY. 

arbitrary,  and  we  may  assume  for  it  such  a  value  as  will  most  sim- 
plify our  calculations.  This  value  is  unity;  that  is,  the  linear  unit 
employed  in  expressing  the  sides  of  our  triangles,  or  other  lines 
considered.  This  value  will  be  generally  used  throughout  this 
treatise. 

9.  To  find  the   length  of  an  arc  of  a  given  number  of  degrees, 
minutes,  etc. 

The  semi-circumference  of  a  circle  whose  radius  is  unity  is  known 
to  be  3*14159265  ;  or,  the  radius  being  R,  the  serni-circurnfereuce  is 
3-14159265  R.  Hence 

When  R  =  1 

Arc  180°  =  3-14159265  R         =  3-141592C5 
«      1°    :=  0-01 7453293  .R        =0-017453293 
"      1'    =  0-0002908882  R     =0-0002908882 
«      1"  =  0-000004848137 /£  =  0-000004848137 

An  arc  x  therefore,  in  the  circle  whose  radius  is  unity,  being  ex- 
pressed in  degrees,  or  minutes,  or  seconds,  we  find  its  length  by  the 
formulae 

Arc  x  =  0-01 7453293  x° 
=  0-0002908882  x' 
=  0-000004848 137  x". 

As  these  factors  for  finding  the  length  of  an  arc  are  often  used,  it 
is  convenient  to  have  their  logarithms  prepared.*  Thus 

Arc  x  =  [8-2418774]  z° 

-  [6-4637261]  x' 

=  [4-6855749]  x" 

in  which  the  rectangular  brackets  are  used  to  express  that  the  logar- 
ithm of  the  factor  is  given  instead  of  the  factor  itself. 

EXAMPLE.  What  is  the  length  of  the  arc  x  =  38°  17'  48",  the 
radius  being  =  1. 

38°  17' 48"  =-137868"  log.     5-1394635 

Log.  factor  for  seconds     4-6855749 

z  =  0-6684031  log.  a;  ^9T825()384 

10.  To  find  the  number  of  degrees,   etc.   in   an   arc   equal  to  the 
radius. 

We  have,  from  the  preceding  article, 


*  The  logarithms  in  the  examples  of  this  work  will  he  taken  from  Stanley's  Tables, 
(published  in  New  Haven,  by  Dnrrie  and  Peck,)  the  best  tables  of  seven-figure  logar- 
ithms yet  published  in  this  country. 


MEASURES  OF  ARCS.  13 

1  80° 


3265 
=  3437'-74677  =  206264'  -806 

11.  The  angle  at  the  center  measured  by  an  arc  equal  to  the  radius,  is  often  taken 
as  the  unit  of  angular  measure,  as  this  angle  will  be  of  an  invariable  magnitude, 
whatever  is  the  length  of  the  radius.  If  x  is  the  number  of  such  units  in  a  given 
angle,  the  number  of  degrees,  etc.,  in  it  will  be  found  by  multiplying  by  the  value  of 
the  radius  in  degrees,  etc.,  found  in  the  preceding  article.  Thus, 

x°  =xR°  =  57°-2957795  z=  [17581226]  x 
x'  =  xR'  —  3437A74677  x  =  [3-5362739]  x 
x"  =  x  R"  —  206264/A806  x  —  [5-3144251]  z 

Reciprocally,  the  angle  being  given  in  degrees,  etc.,  we  reduce  it  to  the  unit  radius, 
by  dividing  by  R°,  R',  or  R",  thus, 


which  is  evidently  the  same  as  multiplying  by  the  factors  of  Art.  9. 

It  appears,  then,  that  an  angle  is  expressed  in  the  unit  of  this  article  by  the  length 
of  the  arc  which  meavsures  the  angle  in  the  circle  whose  radius  is  unity.  Hence,  an 
angle  thus  expressed  is  said  to  be  given  in  arc.  If  we  put  (as  is  usual) 

TT=:  3-14159265-  •  • 

TT  is  the  circular  measure  of  two  right  angles,  or  it  is  the  expression  of  two  right  angles 
in  arc.     In  trigonometry  it  is  therefore  common  to  employ  it  to  denote  an  angular 

magnitude  of  180° ;   —  a  right  angle ;  2  TT  four  right  angles,  etc. 

m 

12.  The  complement  of  an  angle  or  arc  is  the  remainder  obtained 
by  subtracting  the  angle  or  arc  from  90°. 

The  supplement  of  an  angle  or  arc  is  the  remainder  obtained  by 
subtracting  the  angle  or  arc  from  180°. 

Thus  the  complement  of  30°  is  60°  ;  the  supplement  of  30°  is 
150°. 

Two  angles  or  arcs  are  complements  of  each  other  when  their 
sum  is  90°.  They  are  supplements  of  each  other  when  their  sum  is 
180°. 

13.  According   to   these   definitions,  the   complement  of  an   arc 
that  exceeds    90°   is  negative.      Thus  the   complement  of  120°  is 
90°  —  120°  =  —  30°.      In  like  manner  the  supplement  of  200°  is 
180°  — 200°-=  — 20°. 


CHAPTER    II. 

SINES,  TANGENTS,  AND  SECANTS.    FUNDAMENTAL  FORMULAE. 

14.  HAVING  expressed  the  sides  and  angles  of  triangles  by  num- 
bers, we  are  next  to  find  such  relations  between  them  as  shall  enable 
us  to  combine  these  two  different  species  of  quantity  in  computa- 
tion. 

As  every  oblique  triangle  may  be  resolved  into  two  right  triangles 
by  dropping  a  perpendicular  from  one  of  the  angles  upon  the  oppo- 
site side,  the  solution  of  all  triangles  is  readily  made  to  depend  upon 
FIG.  4.  that  of  right  triangles.     Let  us  therefore 

consider  a  series  of  right  triangles,  AS  C, 
A  B'  C',  A  B"  C",  etc.,  Fig.  4,  which  have 
a  common  angle  A.  The  angles  at  Jj, 
B'j  B",  being  also  equal,  the  triangles  are 
similar  ;  and  by  geometry 

BC-.AB  =  B'Cf  '.AB'  =  B"C"  :  AB" 


or  by  the  definitions  of  ratio  and  proportion, 

BC  =  B'C'  =  B"C" 
AB  ~  AB'  ~~~  AB" 

In  like  manner  it  follows  that 

BC      B'C'       B"C" 


and 


AC      AC'     '  AC' 
AB       AB'        AB" 


AC      AC'       AC' 


Hence  it  appears  that  the  ratios  of  the  sides  to  each  other  are  the 
same  in  all  right  triangles  having  the  same  acute  angle  ;  and,  therefore, 
if  these  ratios  are  known  in  any  one  of  these  triangles,  they  will  be 
known  in  all  of  them. 

These  ratios,  then,  depending  on  the  value  of  the  angle  alone,  with- 
out regard  to  the  absolute  lengths  of  the  sides,  may  be  considered  as 
indices  of  the  angle,  and  have  received  special  names,  as  follows : 

15.  The  SINE  of  the  angle  is  the  quotient  of  the  opposite  side  divided, 
by  the  hypotenuse. 

14 


SINES,  TANGENTS,  AND  SECANTS.  15 

Thus,   in    the    right   triangle   ABC,   Fig.   5,  FIG. 5. 

if  we  designate  the  sides  by  the  small  letters 
a,  6,  c,  we  shall  have,  (whatever  the  absolute 
length  of  the  sides) 

a        .     n      b 

sm  A  =  -)     sm  Jo  =  - 
c  c 

16.  The  TANGENT  of  the  angle  is  the  quotient  of  the  opposite  side 
divided  by  the  adjacent  side. 

rr,  .       a  „      b 

Inus  tan.4  =  ->     taui>— - 

6  a 

17.  The  SECANT  of  the  angle  is  the  quotient  of  the  hypotenuse  divided 
by  the  adjacent  side. 

Thus  secJ.=  7>     860.6=:- 

6  a 

18.  The  COSINE,  COTANGENT,  and  COSECANT  of  an  angle,  are  re- 
spectively the  SINE,  TANGENT,  and  SECANT  of  the  complement  of  the 
angle. 

Since  the  sum  of  the  two  acute  angles  of  a  right  triangle  is  one 
right  angle,  or  90°,  they  are,  by  Art.  12,  complements  of  each  other; 
therefore,  according  to  the  preceding  definitions,  we  shall  have 

sin  A  =     cos  B  =  -  cos  A  =  sin  B  =  - 

c  c 

tan  A  =     cotB=-  cot  A  =  tan  B  —  - 

6  a 

sec  A  =  cosec  -6=7  cosec  A  =  sec  B=- 

b  a 

c  n 

19.  Since  -  is  the  reciprocal  of  ->  it  follows  from  the  first  and  last 

a  c 

of  these  equations,  that  the  sine  and  cosecant  of  the  same  angle  are 
reciprocals ;  and  from  the  other  equations,  also,  that  the  cosine  and 
secant,  the  tangent  and  cotangent  are  reciprocals.  That  is, 


(2) 


or  more  briefly, 

sin  A  cosec  A  =  cos  A  sec  A  —  tan  A  cot  A  =  1  (3) 


Sill  A.  - 

cosec  A 

cosec  A 

sin  A 

po^  A  — 

1 

sec  A 

1 

sec  .4 

cos  A 

tan   A  - 

1 

cot  A 

I 

cot  A 

tan  A 

16 


PLANE  TRIGONOMETKY. 


SINES,  ETC.  OF  ARCS. 

20.  The  sine,  tangent,  and  secant  of  an 
arc  are  respectively  the  sine,  tangent,  and 
secant  of  the  angle  at  the  center  measured 
by  that  arc.  Thus,  Fig.  6, 

7?  C1 
— - 


A'" 


The  sine  of  an  arc,  therefore,  does  not  depend  upon  the  absolute 
length  of  the  arc,  but  upon  the  ratio  of  the  arc  to  the  whole  circum- 
ference, (Art.  7.)  It  follows  that  the  relations  (2)  and  (3)  are  also 
applicable  when  A  expresses  an  arc. 

21.  If  the  radius  =  1,  all  the  trigonometric  functions  above  de- 
fined may  be  represented  in  or  about  the  circle  by  straight  lines. 
Representing  the  arc  AB,  or  angle  A  OB,  by  x,  we  have,  when  OA 
=  OB=1, 

BC     EC      „„ 

=  T 


AT     AT 

=~=- 


OT      OT 

sec  x  =  —  —  =     ~  = 
OA         1 


and  from  the  arc  A'B=  90°  —  x  we  find  in  the  •same  way 

cos  x  =  BD=OC 


cosecz—  OT' 

Therefore,  in  the  circle  whose  radius  is  unity,  the  sine  of  an  arc, 
or  of  the  angle  at  the  center  measured  by  that  arc,  is  the  perpendicular 
let  Jail  from  one  extremity  of  the  arc  upon  the  diameter  passing  through 
the  other  extremity. 

The  trigonometric  tangent  is  that  part  of  the  tangent  drawn  at  one 
extremity  of  the.  arc,  which  is  intercepted  between  that  extremity  and  the 
diameter  (produced)  passing  through  the  other  extremity. 

The  secant  is  that  part  of  the  produced  diameter  which  is  intercepted 
between  the  center  and  the  tangent. 

The  cosine  is  the  distance  from  the  center  to  the  foot  of  (he  sine. 

In   a   circle   of  any  other  radius   than   unity,  the  trigonometric 


FUNDAMENTAL  FORMULAE.  17 

functions  of  an  arc  will  be  equal  to  the  lines  drawn  as  above,  divided 
by  that  radius. 

The  properties  here  stated  have  heretofore  been  used  by  most 
writers  upon  trigonometry  as  definitions,  but  without  limiting  the 
radius  to  unity ;  and  it  is  evidently  from  this  mode  of  viewing  these 
functions  that  they  have  derived  their  names. 

22.  Besides  the  functions  already  denned,  others  have  been  occasionally  employed 
to  facilitate  particular  calculations,  as  the  versed  sine,  which  in  the  circle  is  the  portion 
of  the  diameter  intercepted  between  the  extremity  of  the  arc  and  the  foot  of  the  sine ; 
thus,  Fig.  6,  the  versed  sine  of  A  B  is  A  C,  or  the  radius  being  =  1, 

versinz  =  l — cos  a;  (4) 

by  means  of  which  formula  we  may  always  substitute  versed  sines  for  cosines,  and 
reciprocally. 

The  coversed  sine  (covers.)  is  the  versed  sine  of  the  complement,  and  suversed  sine 
(suvers.)  is  the  versed  sine  of  the  supplement. 

The  chords  of  arcs  have  also  been  used,  and  may  be  substituted  for  sines  by  the 
formula 

ch  x  =  2  sin  J  x  (5) 

which  is  evident  from  Fig.  6,  where  if  the  arc  B  B'  =  x,  we  have  chord  B  B'  — 
2  B  C  =  2  sin  A  B. 

23.  From  what  has  now  been  stated,  the  student  will  perceive  that 
angles  are  to  be  subjected  to  computation  by  means  of  the  quanti- 
ties  sine,  cosine,  etc.,  commonly  designated    by  the  comprehensive 
term  trigonometric  functions*     It  becomes  necessary,  therefore,  for 
the  computer  to  know  the  values  of  these  functions  for  any  given 
value  of  the  angle.      The  trigonometric  tables  contain  these  values 
for  every  minute,  and  sometimes  for  every  second,  from  0°  to  90°  ; 
and  with  these  tables  all  the  numerical  computations  of  trigonometry 
are  carried  on.     In  practice,  then,  we  are  not  required  to  compute 
the  functions  themselves,  and  we  shall  therefore  defer  the  methods 
for  that  purpose  to   a  subsequent  part  of  this  work,  and  proceed 
at  once  with  the  investigation  of  the  formulae  and  methods  by  which 
these  tables  are  rendered  available. 

FUNDAMENTAL  FORMULAE. 

24.  Given  the  sine  of  an  angle,  to  find  the  cosine. 

From    the    right    triangle    ABC,   Fig:  7,  we  FlG-7- 

have  by  geometry         a2  4-  b2  —  c2 
Dividing  by  c2,  this  equation  becomes 


*  Also  trigonometric  lines,  from  the  properties  explained  in  Art.  21. 


18 


PLANE  TRIGONOMETRY. 


or,  by  the  definitions  of  sine  and  cosine  (1), 

sin2  A  +  cos2  A  =  1  (6) 

in  which  the  notation  sin2  A  signifies  "  the  square  of  the  sine  of  A." 
From  this  formula,  if  the  sine  is  given,  we  find 

cos2  A  =  1  —  sin2  A  =  (1  +  sin  A)  (1  —  sin  A) 
cosA=y(l  —  sin2  A)  =  y  [(1  +  sin  A)  (I  —  sin  A)]         (7) 
and  if  the  cosine  is  given,  we  find 

sin  A=y(l  —  cos2  A)=y  [(1  +  cos  A)  (1  -  cos  A)]        (8) 
25.   Given  the  sine  and  cosine  of  an  angle,  to  find  the  tangent. 
By  (1)  we  have 


also 


therefore 


tan  A  ==- 
b 


sin  A 
cos  A 

tan  A  = 


c       c 
sin  A 


cos  A 

And  since  the  cotangent  is  the  reciprocal  of  the  tangent, 

cos  A 


cot  A  = 


sin  A 


(9) 


(10) 


26.   Given  the  tangent  of  an  angle,  to  find  the  secant. 
The  right  triangle  A  B  C,  Fig.  7,  gives 

c»=62  +  aa 
Dividing  by  b2,  this  becomes 

r2  nz 

-=14-- 

62          r  62 

or,  by  the  definitions  of  secant  and  tangent  (1), 

sec2  A  =  1  +  tan2  A 

This  formula  applied  to  the  complement  of  A  gives 
cosec2  A  =  1  +  cot2  A 

27.  The  preceding  formulae  are  also 
directly  obtained  from  Fig.  8.  If  the 
angle  A  O  R,  or  the  arc  A  B,  be  denoted 
by  x,  the  right  triangle  OJ3C,  gives 


(11) 
(12) 


or  remembering  that  the  radius  is  unity, 
by  Art.  21, 
sin2  x  +  cos2  x  =  1  (13) 


FUNDAMENTAL  FORMULAE.  19 

The  triangle  OB  Ogives  by  the  definition,  Art.  16, 


sn  a;  -   . 

or  tana;-  (14) 

cos  a; 

Since  the  angle  BOD  is  the  complement  of  JBOC,  tan  BOD  = 
cot  a*,  and  the  triangle  BOD  gives 

BD       00 


cos  a;  /t   N 

or                                                 cot  x  =-  —  —  (15) 

sin  x 

In  a  similar  manner  the  triangles  A0rl\  A'  OT'  give 

sec2  a;  —  1  -f  tan2  x  (16) 

cosec2  x  —  1  +  cot2  x  (17) 

28.    The  following   equations   are   easily   demonstrated    by  combining   (13),   (14), 

(15),   ^16),  (17),  and  employing  the  property  of  the  reciprocals   (2).  They  are  of 
frequent  use. 


sin  z  =  • 


cosec  z  sec  x       cot  z 

1  •              cot  z        sin  z                                     /1nx 

• -=cotzsinx  =  —  (19) 

sec  z  cosec  z      tan  z 

sin  z  =  i/  (1  —  cos1  z),  cos  z  =  -j/  (1  —  sin1  z)                             (20) 

sec  z  —  i/  ( 1  -f  tan2  z),  cosec  z  =  -j/  (1  +  cot1  z)                              (21 ) 

(sec'z  — 1),  cot  x  =  l/  (cosec2  z  — 1)                          (22) 

tan  z  1 


I/  (1  +  tan1  z)       i/  ( 1  +  cot2  z) 


(23) 


«»*=7T7T^~7rr=        „    ,V  ,  .  (24) 

(25) 

(26) 
\/  (1  —  cos2  z)  sinz 

29.   To  find  the  sine,  etc.  of  30°  and  60°. 

In  Fig.  8,  let  the  arc  A  B  =  30°,  and  BB'  =  2AB  =  60°.  By 
Art.  21,  sin  AB  =  BO,  and  by  geometry  the  chord  of  60°,  or  of  one- 
sixth  of  the  circumference,  is  equal  to  the  radius  —  1  ;  therefore 

2  sin  30°=-  2  BC=^  BB'  =  1 
whence  sin  30°  =  I  =  cos  60°  (27) 


20  PLANE  TRIGONOMETRY. 

and  by  (7) 

cos  30°  =  y  [(1  +  J)  (1  -  1)]  =  „( 
whence  cos  30°  =  £  y  3  =  sin  60° 

Then,  by  (9)  and  (2), 


X  J) 


-  —  cot  60° 


sin  30°          \  1 

tan  30°  =  -  — 

cos  30°      \ 

cot  30°  =  -        -  =  r/  3  =  tan  60° 

tan  30° 

sec  30°  =  -  =  —  =  cosec  60° 

cos  30°       ]/3 


1 


=  2  =  sec  60C 


(28) 

(29) 
(30) 
(31) 
(32) 


cosec  30°  = 

sin  30° 

30.   To  find  the  sine,  etc.  of  45°.     Since  45°  is  the  complement  of 
45°,  we  have 

sin  45°  =  cos  45° 
whence  by  (13),  putting  x  =  45°, 

sin2  45°  +  cos2  45°  =  2  sin2  45°  =  2  cos2  45°  =  1 

sin2  45°  =  cos2  45°  —  £ 
sin  45°  =  cos  45°  =  V\  =  \  1/2  (33) 


tan  45°  -  cot  45°  == 


sec  45°  -    cosec  45°  = 


cos  45° 
1 


=  1 


sin  45° 


(34) 
(35) 


These  values  are  readily  verified  in  the  circle, 
Fig.  9,  where  OAT  A'  is  a  square  described  upon 
the  radius.  The  diagonal  0 T  bisects  the  right 
angle,  whence  AOT=  45°,  and  tan  45°  =  A  T 

=  OA  =  1 ;  cot  45°  =  A'T=l;  sin  45°  =  BC 

=  OC=cos45°,  etc. 

31.  The  sines  and  cosines  of  two  angles  being  given,  to  find  the 
sine  and  cosine  of  the  sum,  and  the  sine  and  cosine  of  the  difference 
of  those  angles. 

Fl°-  »•  Let  the  two  angles  be  A  0  B 

and  BOC,  Figs.  10  and  11. 
At  any  point  B  in  the  line 
OB  draw  BC  perp.  to  OB. 
Draw  BA  and  CD  perp.  to 
OA,  and  7?  A1  perp.  to  CD. 


FUNDAMENTAL  FORMULA.  21 

Theii  the  triangles  BCE  and  BOA  are  mutually  equiangular, 
the  three  sides  of  the  one  being  perp.  to  the  three  sides  of  the  other 
respectively  ;  therefore  the  angle  B  C  E  =  A  0  B. 

Let  x  =  AOB  =  BCE 

y  =  BOC 

Then,  Fig.  10,  x  +  y  =  COD 

Fig.  11,  x  —  y  =  COD 

and  in 

,       CD      BA  +  CE      BA.CE 
O,8m(*  +  3,)==-  =     --      - 


.  _  CD       BA  -  CE       BA      CE 

1,31  -y):-  —  -         ~CQ  =  -CO~w 

and  in  both  figures 

BA      BA  ^  BO 

cd  =  Bdxco^^<^ 

CE      CE       CB 


which  being  substituted  in  the  above  expressions  of  sin  (x  +  y)  and 
sin  (x  —  y)  give 

sin  (a;  -f-  y)  =  sin  x  cos  y  -(-  cos  x  sin  y  (36) 

sin  (a;  —  y)  =  sin  x  cos  y  —  cos  x  sin  y  (37) 

Again  in 

„.  OD      OA-EB      OA      EB 

F.g.10,         oos(*  +  y 


Fig.  11,  008^-^ 

and  in  both  figures, 

OA       OA       OB 


07)       OA  +  EB      OA    .   EB 


EB      EB    ,  BC 
OC  =  BCXOC=*}nX*my 
therefore 

cos  (x  -|-  y)  —  cos  x  cos  ?/  —  sin  x  sin  y  (38) 

cos  (re  —  y)  =  cos  x  cos  y  +  sin  x  sin  y  (39) 

and  (36),  (37),  (38),  and  (39)  are  the  required  formulae. 

These  may  be  considered  as  the  fundamental  formulae  of  the  trigo- 
nometric analysis,  and  will  form  the  basis  of  our  subsequent  inves- 
tigations. They  are  equally  applicable  to  arcs  represented  by  x  and 
y  (Art.  20). 


CHAPTER    III. 


TRIGONOMETRIC  FUNCTIONS  OF  ANGULAR  MAGNITUDE  IN 

GENERAL. 

32.  THE  definitions  of  sine,  etc.  given  in  the  preceding  chapter 
apply  only  to  acute  angles,  since  the  angle  is  there  assumed  to  be  one 
of  the  oblique  angles  of  a  right  triangle.     But  we  shall  now  take  a 
more  general  view  of  angular  magnitude  and  of  the  functions  by 
means  of  which  it  is  subjected  to  computation. 

If,  Fig.  12,  we  suppose  the  line  OA  to  revolve 
from  the  position  OA  to  OA'  in  the  direction  of 
the  arc  AA'  (or  from  right  to  left),  it  will  describe 
an  angular  magnitude  of  90°  ;  when  it  arrives  at 
OA"  it  will  have  described  an  angular  magnitude 
of  180°  ;  at  OA'",  270°;  and  at  OA  again,  360°. 
If  it  now  continue  its  revolution,  when  it  arrives 
at  OA'  again,  it  will  have  described  an  angular  magnitude  of  360° 
+  90°,  or  450°  ;  and  thus  we  may  readily  conceive  of  an  angular 
magnitude  of  any  number  of  degrees.  In  like  manner  we  may  have 
arcs  equal  to  or  greater  than  one,  two,  or  more  circumferences. 

To  obtain  trigonometric  functions  for  angles  and  arcs  thus  gene- 
rally considered,  we  shall  avail  ourselves  of  the  fundamental  formu- 
lae established  in  the  preceding  chapter  ;  first  deducing  their  values 
analytically,  and  then  explaining  their  geometrical  signification. 

33.  To  find  the  sine,  etc.  of  0°  and  90°.    In  (37)  and  (39)  let  x  =  y  ; 
the  first  members  become  sin  (x  —  #)  =  sin00,  and  cos  (x  —  x)  =  cos 
0°  ;  and  by  (13)  they  are  reduced  to 

sin  0°  =  sin  x  cos  x  —  cos  x  sin  x  =  0 
cos  0°  =  cos2  x  -+-  sin2  x  =1 

and  since  0°  and  90°  are  complements  of  each  other,  Art.  12, 
sin  0°  =  cos  90°  =  0 
cos  0°  =  sin  90°  =  1 
from  which  by  (9)  and  (2) 


tan  0°  =  cot  90° 


snO     _^0 

cos  0°  ~  1  ~ 


(40) 
(41) 

(42) 


22 


FUNCTIONS  OF  ANGULAK  MAGNITUDE.  23 

cotO°=     *»'90°=-i-=l=»  (43) 

sec  0°  =  cosec  90°  =  -              \  =  l  (44) 
cos  0         1 

cosec  0°=     sec  90°  =  -      -=^  =  00  (45) 
sin  0         0 

34.  To  find  the  sine,  etc.  of  180°.  In  (36)  and  (38)  let 
x  =  y  —  90°  ;  these  equations  become  by  means  of  the  preceding 
values 

sin  180°  =  1  X  0  +  0X1  —  0  (46) 

cos  180°  =  0X0-1X1  =  —  !  (47) 
whence  by  (9)  and  (2) 

tan  180°=           -0                   cot  180°=       -  <x>  (48) 
—  1                                                   U 

sec  180°  =  -^-  =  —  1           cosec  180°  =  -  =  oo  (49) 


35.  To  find  the  sine,  etc.  of  270°.     In  (36)  and  (38)  let  x  =  180°, 
y  =  90°,  then 

sin  270°  —  OXO  +  (—  1)  X  1=      -1  (50) 

cos270°  =  (-l)XO—  OX  1  =  0  (51) 

tan  270°  =  —  -  =  oo  cot  270°  =  —  =  0  (52) 

0  oo 

sec  270°  =-  =  oo  cosec  270°  =  -j-  =     -1   (53) 

36.  To  find  the  sine,  etc.   of  360°.      In  (36)  and  (38)  let  x  =  y 
=  180°;  then 

sin360°  =  OX(-l)  +  (—  1)XO  =  0  (54) 

cos  360°  =  (—  1)  x  (—  1)  -0X0=1  (55) 

the  same  values  as  for  0°,  whence  it  follows  that  all  the  trig,  func- 
tions of  360°  are  the  same  as  those  of  0°. 

The  same  process  continued  will  give  for  450°  (  =  360°  +  90°), 
the  same  trig,  functions  as  those  of  90°  ;  fer  540°  the  same  functions 
as  for  180°,  etc. 


24  PLANE  TRIGONOMETRY. 

37.  The  preceding  values  now  furnish  us  at  once  with  the  values 
of  the  functions  for  all  possible  values  of  the  angle.     In  (36)  and 
(38)  let  x  —  00,  they  are  reduced  to 

sin  y  =  sin  0°  cos  y  +  cos  0°  sin  y  =  sin  y 
cos  y  =  cos  0°  cos  y  —  sin  0°  sin  y  =  cos  y 

which  are  simply  identical  equations,  and  reveal  no  new  property. 
But  if  in  (37)  and  (39)  we  put  x  =  0°,  we  have,  after  substituting 
the  functions  of  0°, 

sin  ( —  y)  =  —  siny  cos  ( — y)  =  cosy  (56) 

whence  by  (9)  and  (2) 

(57) 
(58) 
(59) 

(60) 

or,  the  sin.,  tan.,  cot.,  and  cosec.  of  the  negative  of  an  angle  are  the 
negative  of  the  sin.,  tan.,  cot.,  and  cosec.  of  the  angle  itself ;  and  the  cos. 
and  see.  are  the  same  as  those  of  the  angle  itself. 

38.  In  (37)  and  (39)  let  .r  =  90°;  we  find  after  reduction 

sin  (90°  -—  y)  —  cos  y          cos  (90°  -—y)  =  sin  y 

which  agree  with  the  definition  of  cosine,  but  give  no  new  relations. 
But  in  (36)  and  (38)  let  x  =  90°,  we  find 

sin  (90°  +  y)  =  cos  y,          cos  (90° -f  y)  =    -  sin  y          (61) 
whence  by  (9)  and  (2), 

tan  (90°  +  y)  =  —  cot  y  cot(90°  +  .y)  =    -  tan  y         (62) 

sec  (90°  +  y)  =  —  cosec  y          cosec  (90°  +  y)  =  sec  y         (63) 

or,  the  sin.  and  cosec.  of  an  angle  are  equal  to  the  cos.  and  sec.  of  the 
excess  of  the  angle  above  90°  ;  and  the  cos.,  tan.,  cot.,  and  sec.  are 
equal  to  the  negatives  of  the  sin.,  cot.,  tan.,  and  cosec.  of  the  excess  of 
the  angle  above  90°. 


sin(  —  y)_ 

\. 
—  sin  y 

tan  ^      y) 

cos(—  y) 
cos(  —  y) 

cosy 
cos  ?/ 

sin(—  y) 
1 

1 

cot  i/ 
ecy 
-  cosec 

sec  (      y) 
jsec  (  —  y)  = 

cos(—  y) 

1 

cosy 

1 

sin(—  y) 

—  siny 

FUNCTIONS  OF  ANGULAR  MAGNITUDE.  25 

30.  In  (37)  and  (39)  let*  =  180°  ;  we  find 

sin  (180°  —  y)  =  sin  y  cos  (180°  —  y)  =    -  cos  y  (64) 

tan  (180°  —  y)=    -  tan  y  cot  (1 80°  —  y)  =  —  cot  y  (65) 

sec  (180°  —  y)  =  —  secy  cosec  (180°  —  y)  —  cosec  y  (66) 

or,  the  sin.  and  cosec.  of  the  supplement  of  an  angle  are  the  same  as 
those  of  the  angle  itself;  and  the  cos.,  tan.,  cot.,  and  sec.  are  the  nega- 
tive of  those  of  the  angle  itself. 

40.  If  y  is  acute  (that  is,  less  than  90°),  all  its  trig,  functions  are 
positive;  and  since  its  supplement  180°  — y  is  obtuse  (that  is,  greater 
than  90°),  it  follows  from  the  preceding  article,  that  the  sin.  and  cosec. 
of  an  obtuse  angle  are  positive)  while  its  cos.,  tan.,  cot.,  and  sec.  are 
negative. 

41.  In  (36)  and  (38)  let  x  =  180°  ;  we  find 

sin  (180°  +  y)  =    -  sin  y  cos  (180°  +  y)  =  -  cos  y      (67) 

tan  (180°  +  y)  =  tan  y  cot  (180°  +  y)  =  cot  y  (68) 

sec  (180°  +  y)  =    -  sec  y         cosec  (180°  +  y)  =    -  cosec  y  (69) 

by  means  of  which,  if  y  is  acute,  we  obtain  the  values  of  the  sines, 
etc.  of  angles  between  180°  and  270°. 

42.  In  (37)  and  (39)  let  x  =  270°  ;  we  find 

sin  (270°  —  y)  =  —  cos  y  cos  (270°  —  y)  =  —  sin  y      (70) 

tan  (270°  —  y)  ==  cot  y  cot .(270°  —  y)  =  tan  y  (71) 

sec  (270°  —  y)  =  —  cosec  y      cosec  (270°  —  y)  =  —  sec  y      (72) 

43.  In  (36)  and  (38)  let  x  =  270°  ;  we  find 

sin  (270°  +  y)  =  —  cos  y  cos  (270°  +  y)  =  sin  y  (73) 

tan(270°  +  y)=    -  cot  y  cot  (270°  +  y)  =  —  tan  y      (74) 

sec  (270°  +  y)  =  cosec  y          cosec  (270°  -f  y)  =  —  sec  y       (76) 

44.  In  (37)  and  (39)  let  x  =  360°  ;  we  find 

sin  (360°  —  y)  =    -  sin  y  cos  (360°  —  y)  =  cos  y  (76) 

tan  (360°  —  y)  =  —  tan  y  cot  (360°  —  y)  =  —  cot  y       (77) 

sec  (360°  —  y)  =  sec  y  cosec  (360°  —  y)  =  —  cosec  y   (78) 

or  the  functions  of  360°  —  y  are  the  same  as  those  of —  y  (Art.  37). 

45.  In  (36)  and  (38)  let  x  =  360°  ;  we  find 

sin  (360°  +  y)  =  sin  y  cos  (360°  +  y)  =  cos  y  (79) 

or,  the  functions  of  an  angle  which  exceeds  360°  are  the  same  as  those 
of  the  excess  above  360°. 

4  C 


26 


PLANE  TRIGONOMETRY. 


It  follows  that  the  functions  of  720°  -J-  y  are  the  same  as  tho.se  of 
360°  -j-  y,  and  therefore  the  same  as  those  of  y  ;  and  in  like  manner 
for  an  angle  which  exceeds  any  multiple  of  360°. 

46.  Since  .y  —  90°  is  the  negative  of  90°  --y,  we  obtain  from 
Art.  37, 

sin  (y-  90°)=    -  sin  (90°  -  y)  =    -cosy  ) 

cos  (y  —  90°)  =  cos  (90°  —  y)  =  sin  y  C  ' 

whence  also  tan.,  etc.  ;  and  in  the  same  manner  we  may  find  the  func- 
tions of  y  —  180°,  y  —  270°,  y  —  360°,  etc. 

47.  We  shall  now  give  the  geometrical  interpretation  of  the  pre- 
ceding results. 

In  Fig.  13,  let  the  radius  revolve  from  the 
position  OA  to  OA',  OA",  etc.,  as  in  Art.  32, 
thus  describing  a  continuously  increasing  an- 
gular magnitude  ;  or,  which  is  equivalent,  let 
the  arc   commencing    at    A    increase   contin- 
uously  to   AB,  A  A',   AB',   etc.      Then   the 
changes  in  the  values  of  the  several  trigono- 
metric lines  may  be  traced  as  follows. 
1st.   The  sine  being,  by  Art.  21,  the  perpendicular  from  one  extre- 
mity of  the  arc  upon  the  diameter  drawn  through  the  other  extremity, 
we  shall  have  sin  AB  =  BC,  sin  AB'  =  B'  Cf,  sin  A  A"  B"  =  B"C', 
sin  AA"  B'"  =  B'"  C,  and  if  we  make 

AB  =  A"  B'  =  A"  B"  =  AB'"  =  y 
we  have 

sin  y  =  B  C 

sin  (180°  -y)  =  B'  C' 


B'" 


sin  (360°  —  y)  =  B'"  C 

The  lines  BC,  B'C',  B"C',  Bf"  C,  however,  represent  only  the 
numerical  values  of  the  sines,  and  are  here  equal.  But  the  results 
above  obtained  from  our  formulae  enable  us  to  distinguish  between 
them  by  means  of  their  algebraic  signs.  Thus,  by  (64),  (67),  (76), 

sin  (180°  —  y)  =  siny 
sin  (180°  +  y)=    -siny 
sin  (360°  —  y)  =    -  sin  y 

so  that  the  sines  from  0°  to  180°  are  positive,  while  those  from  180° 
to  360°  are  negative  ;  or  the  sines  which  are  above  the  diameter 
A  A"  are  positive,  while  those  which  are  below  this  diameter  are 


FUNCTIONS  OF  ANGULAR  MAGNITUDE. 


27 


negative;  or  still  more  generally,  the  sines  that  have  opposite  di- 
rections, with  reference  to  the  fixed  diameter  from  which  they  are 
measured,  have  opposite  signs. 

2d.   The  cosine  being,  by  Art.  21,  the  distance  from  the  center  to 
the  foot  of  the  sine,  we  have 

cos  y  =  O  C 
cos(180°  —  2/)  = 


cos  (360°  —  y)  =  OO 
but  by  (64),  (67),  (76), 

cos  (180°  —  y)  =  —  cos  y 
cos  (180°  +  y)=  —  cos  y 
cos  (360°  —  y)  =  cos  y 

so  that  the  cosines  on  the  right  of  the  diameter  A'  A'"  are  positive, 
while  those  on  the  left  of  this  diameter  are  negative  ;  or  rather  the 
cosines  that  have  opposite  directions,  with  reference  to  the  diameter 
from  which  they  are  measured,  have  opposite  signs. 

We  have  here  only  exhibited  a  well-known  principle  in  the  appli- 
cation of  analysis  to  geometry,  viz.  :  that  all  lines  measured  in  opposite 
directions  from  a  fixed  line  have  opposite  signs. 

To  interpret  the  results  (56),  it  is  only  necessary  to  observe  that 
a  negative  arc  will  be  one  reckoned  from  A  towards  B"1  ',  or  in  the 
opposite  direction  to  that  of  the  positive  arc,  so  that 

sin  A  B'"  =  sm(—y)  =  B"r  C  =  -  B  C=  —  sin  y 
cos  A  B'"  —  cos  (  —  y)  =  0  C=  cos  y 

as  in  (56). 

The  same  principle  applies  to  the  tangents,  but  it  will  be  simpler 
in  practice  to  obtain  their  signs  (as  also  those  of  the  secants),  ana- 
lytically, from  those  of  the  sine  and  cosine,  as  has  been  already  shown. 
It  will  be  sufficient  to  bear  in  mind  the  following  table,  which  is  also 
expressed  by  Fig.  13. 


SINE 
COSINE 

1st  QUAD. 

2d  QUAD. 

3d  QUAD. 

4th  QUAD. 

+ 

+ 

— 

+ 

28 


PLANE  TRIGONOMETRY. 


48.  The  particular  values  of  the  sine  and 
cosine  at  A,  Ar,  A",  etc.,  or  sin.  and  cos.  of 
0°,  90°,  180°,  etc.,  may  also  be  found  by  Fig. 
13,  upon  the  same  principles ;  but  this  we  leave 
to  the  student. 

49.  GENERAL   REMARK. — In  the   demon- 
stration of  the  fundamental  formulae   for  sin 
(xrti/),  and  cos  (x±y\  Art.  31,  the  angles 

x,  y  and  x  ±y  were  all  taken  less  than  90°  and  positive.  In  this 
chapter  these  formulae  have  been  applied  to  angles  of  any  magnitude, 
and  the  resulting  functions  have  been  shown  to  take  opposite  signs 
when  the  lines  representing  them  take  opposite  directions.  It  follows 
that,  in  deducing  trigonometric  formulae  from  geometrical  figures,  we 
need  not  embarrass  our  demonstrations  with  the  consideration  of  the 
various  cases  of  the  problem,  or  of  the  various  values  of  the  angles 
of  the  figure.  The  formula  deduced  from  any  supposed  position  of 
the  lines  of  the  figure  will  be  of  general  application,  provided  in  the 
practical  application  of  this  formula  to  the  particular  cases,  we  observe 
those  values  and  signs  of  the  trigonometric  functions  which  have  now 
been  determined. 


50.  The  results  of  this  chapter  may  be  expressed  by  a  few  general  formulae.  From 
(79)  it  appears  that  all  the  trigonometric  functions  return  to  the  same  values  after  one 
or  more  complete  revolutions  of  3fiO°.  If  we*  represent  the  semi-circumference,  or 
two  right  angles,  hy  TT  (Art,  11),  and  let  n  =  any  whole  number  or  zero,  we  shall 
have 


sin  4  n  —  =  0 

sin  (4n  +  1)  —  =1 


sfn  (4  n  -f  2)  —  =  0 

4U 


sin  (4  n  -f-  3) 


— 1 


whence 


tan  4  n  —  —  0 

m 


tan(4n+'2) 


A  ~  1 

cos  4  n  —  =  1 


cos  (4  n  +  2)  —  =  —  1 
35 


cos  (4  n  +  3)  —  =  0 


tan  (4  n  -f-  1 )  —  =  00 


tan  (4  n  -f-  3)  —  =00 

2 


(81) 
(82) 
(83) 
(84) 


or  the  tan.  of  the  even  multiples  of  —  =0,  and  of  the  odd  multiples  =  oo,  so  that  we 

SS 

may  write  more  simply 


tan  2  n  —  =  0 
2 


tan  (2  n  -\-  1)  —  =  00 
z 


(85) 


FUNCTIONS  OF  ANGULAR  MAGNITUDE.  29 

In  these  formulae  we  have  only  to  give  n  one  of  the  values  0,  1,  2,  3,  4,  etc.,  to 
obtain   the   functions   of   any   given   multiple   of   the   right   angle.      Thus,  we   find 

sin  450°  —  sin  5  -  =  sin  (4  4  1)  ^  =  1  by  making  n  —  1,  in  (82). 

Since  the  subtraction  of  8  n  -  from  the  arc  will  not  change  the  functions,  the  above 

2t 

formulae  are  also  true  when  n  is  a  negative  whole  number. 
51.  In  a  similar  manner  we  obtain 

sin  Hi  n  ~  4  yl  =  sin  y  cos  J4  n  ~  4  yj  =  cos  y        (86) 

sin  j   (4  n  4  1)  —  4  y  1  —  cos  y  cos  J(4  n  4  1)  -|-  4  y  \  =  —  sin  y  (87) 

sin  £(4n  +  2)|-4y]=   -  sin  y  cos  [  (4  n  4  2)  -|  4  yl  ==  —  cos  y  (88) 

|—  — i  p-  ~^ 

sin      (4n43)— 4y     =  — cosy  cos      (4w  4  3)  —  4  y     —  Biny        (89) 

L  2         J  L  2         J 

tan  I  2  ?i  —  4  y  I  =  tan  2/  tan  I   (2«.41)—+y|=  —  coty  (90) 


in  which  n  may  be  any  whole  number,  positive  or  negative,  and  y  any  angle,  positive 
or  negative. 

52.  A  still  more  concise  form  may  be  given  to  the  formulje  of  the  two  preceding 
articles,  as  follows:  n  being,  as  before,  any  whole  number,  positive  or  negative. 

sin2n^-:=0  cos  2n-  =  (  —  l)n  (91) 

sin  (2n  +  1)-  =  (  —  !)»  cos  (2n  +  1)  -  =0  (92) 

2t  ** 

=(-l)nsiny  cos  [2  n  |-  +  yj  =  (-!)»  cosy    (93) 

=(-l)»co8y    cos      (2n  +  1)       -\  !/    =  -(-l)»Biny(94) 


and  from  these  (85)  and  (90)  may  be  directly  deduced. 

53.  We  have  seen  that  an  angle  being  given,  there  is  but  one  corresponding  sine. 
On  the  other  hand,  a  sine  being  given,  there  is  an  indefinite  number  of  angles  corre- 
sponding; for  if  a  denote  the  given  sine,  and  y  any  corresponding  angle,  then  o  is  also 
the  sine  of  all  the  angles 

if  —  y,  2  TT  -f  y,  3  IT  —  y,  etc. 

_,r  —  y,  _27r4y,  _  3  n-  —  y,  etc. 

or  in  general 

a  =  siny  =  sin  [n?r  4  (  —  l)"y]  (95) 

c2 


30  PLANE  TRIGONOMETRY. 

In  like  manner  if  a  is  a  given  cosine,  and  y  any  corresponding  angle, 

a  =  cos  y  =  cos  (2  n  TT  ±  y)  (96) 

and  if  a  is  a  given  tangent  corresponding  to  the  angle  y, 

a  =  tan  y  =  tan  (n  if  +  y)  (97) 

SINE  AND  TANGENT  OF  A  SMALL  ANGLE  OR  ARC. 

54.  When  the  angle  A  0  B  =  x,  Fig.  14,  is 
very  small,  the  sine  and  tangent  are  very  nearly 
equal  to  the  arc  A  B,  which  measures  the  angle, 
the  radius  being  unity ;  and  the  cosine  and  secant 
are  nearly  equal  to  OA  =  l  (Art.  21).  There- 
fore, to  find  the  sine  or  tangent  of  a  very  small 
angle  approximately,  we  have  only  to  find  the 
length  of  the  arc  by  Art.  9 ;  thus 

sin  1"  =  arc  I"  =  0000004848137 

log.  sin  1"  =  4-6855749 
and  x  being  a  small  angle,  or  arc,  expressed  in  seconds, 

sin  x  =  tan  x  =  x  sin  1 "  (98) 

If  #  is  expressed  in  minutes, 

sin  x  =  tan  x  —  x  sin  V  (99) 

If  x  expresses  the  length  of  the  arc,  the  radius  being  unity, 

sin  x  =  tan  x  =  x  (1 00) 

The  employment  of  these  approximate  values  must  be  governed  by 
the  degree  of  accuracy  required  in  a  particular  application.  It  is 
found,  for  example,  that  they  are  sufficiently  accurate  when  the 
nearest  second  only  is  required  in  our  results,  provided  the  angle 
does  not  much  exceed  1°. 

55.  If  x  and  y  are  any  two  small  angles,  it  follows  from  the  pre- 
ceding article  that 

sin  x  :  sin  y  =  x  sin  V  :  y  sin  1"  =  x  :  y 

that  is,  the  sines  (or  tangents)  of  small  angles  are  proportional  to  the 
angles  themselves.  The  application  of  this  theorem,  however,  like 
that  of  the  preceding,  must  depend  upon  the  accuracy  required  in  the 
problem  in  which  it  is  employed.* 

*Fora  full  discussion  of  the  limits  under  which  this  theorem  maybe  employed, 
see  a  paper,  by  the  author  of  this  work,  in  the  Astronomical  Journal,  (Cambridge, 
Mass.)  Vol.  i.  p.  84. 


CHAPTER    IV. 

GENERAL  FORMULAE. 

56.  WE  have  already  obtained  four  fundamental  equations,  (36), 
(37),  (38),  and  (39),  involving  two  angles,  x  and  y.     From  these  we 
shall  now  deduce  a  number  of  formula,  either  required  in  the  sub- 
sequent parts  of  this  work,  or  of  general  utility  in  the  applications 
of  trigonometry. 

57.  The  sum  and  difference  of  the  equations  (36)  and  (37)  are 

sin  (x  -f-  y)  -\-  sin  (x  —  y)  —  2  sin  x  cos  y 
sin  (x  +  y]  —  sin  (x  —  y)  —  2  cos  x  sin  y 
and  the  sum  and  difference  of  (38)  and  (39)  are 

cos  (x  -j-  y)  +  cos  (x  —  y)  =  2  cos  x  cos  y  (103) 

cos  (x  -f-  y)  —  cos  (x  —  y)  —    —  2  sin  x  sin  y  (104) 

58.  If  we  put 

x  -f-  y  =  x' 

x  —  y  =  y' 

whence  2  x  =  x'  +  y',  x  =  |  (x'  -f-  y') 

2y  =  x'  —  y',  y=i(x'  —  y') 

equation  (101)  will  become 

sin  x'  +  sin  y'  =  2  sin  1  (#'  +  y'}  cos  i  (#'  —  y') 

and  (102),  (103),  and  (104)  admit  of  a  similar  transformation.  But 
since  x'  and  y'  admit  of  all  varieties  of  value,  we  may  omit  the 
accents  and  apply  the  formulae  to  any  two  angles  x  and  y,  we  have 
thus 

sin  x  -f  sin  y  —  2  sin  |  (x  +  y)  cos  %  (x  —  y)  (1 05) 

sin  x  —  sin  y  =  2  cos  |  (z  -f-  y)  sin  f  (z  —  y)  (106) 

cos  #  +  cos  y  =  2  cos  |  (#  +  y)  cos  |  (x  —  y)  (107) 

cos  x  —  cos  t/  =    —  2  sin  |  (.r  +  T/)  sin  |  (a-  —  2/)  (108) 

Each  of  these  equations  may  be  enunciated  as  a  theorem  ;  thus 

31 


32  PLANE  TRICxONOMETRY. 

(105)  expresses  that  "the  sum  of  the  sines  of  any  two  angles  is 
equal  to  twice  the  sine  of  half  the  sura  of  the  angles  multiplied  by 
the  cosine  of  half  their  difference." 

These  formulae  are  of  frequent  use  (especially  in  computations 
performed  by  logarithms),  in  transforming  a  sum  or  difference  into 
a  product. 

59.  Dividing  (105)  by  (106),  we  have  by  (14)  and  (15) 


or  by  (2) 


sin  x  -f-  sin  y 

— r-*  =  tan  £  (x  +  y)  cot  $  (a;  —  y) 
sin  x  —  sin  y 


sin  x  -f  sin  y  _  tan  i^_+  y} 
sin  x  —  sin  y       tan  J  (x  —  y) 


and  from  (107)  and  (108)  we  find  in  the  same  manner 

-ten^  +  yJtanHs-y)  (HO) 

cos  x  +  cos  y 

We  find  also 

sin  x  -\-  sin  y 

— *  =  tan  \  (x  +  y)  (111) 

cos  x  +  cosy 

sin  x  —  sin?/  N 

-tanifa  —  y)  (112) 

cos  x  +  cos  y 

sin  x  -f  sin  ?/  , 

~eot|(s— y)  (113) 

cos  x  —  cos  y 

sin  a? — sin  y  . 

-cot|(a?-f-y)  (114) 

cos  x  —  cos  y 

60.  Divide  the  equations  (36),  (37),  (38)  and  (39)  by  cos  x  cos  y ; 
then  by  (14)  we  have 

sin  (.T:  -f  ?/) 

J— ^  —  tan  x  +  tan  y  (11  o) 

cos  a;  cos  y 

sin  (a;  —  y} 

=  tanar  — tany  (1 16) 

cos  x  cos  y 

cos  (x  +  ?/)        - 

z-  =  1  —  tan  #  tan  v 
cos  x  cos  y 

cos  (#  —  y) 

•      =  1  +  tan  .7;  tany  (118) 

cos  x  cos  y 


GENERAL  FORMULAE.  33 

61.  Divide  (36),  (37),  (38)  and  (39)  by  sin.r  siny;  and  by  sin  x 
cos  y  ;  then 

sin  (re  ±v) 

v        v>  =coty  ±cot:c 

sin  #  sin  y 

cos  (x  ±  y] 

—r—  v  —  ;-*"-  —  cot  x  cot  y  q=  1  (1  20) 

sin  #  sin  y 


/ioi\ 
1  ±  cot  x  tan  y  (121) 

cos  (#  ±  y)  /.,  oox 

=  cot  cr  :+:  tan  y  (122) 

sin  x  cos  y 

62.  Divide  (115)  by  (117),  and  (110)  by  (118);  then  by  (14) 

»         tan  x  -f-  tan  y  /1  0.,, 

tan  (x  -{-  y)  =  (1  23) 

1  —  tan  x  tan  y 

tana;  —  tan  y  /io,i\ 

tan  (a?  —  y)  =  -  (124) 

1  +  tan  x  tan  y 

by  which,  when  the  tangents  of  two  angles  are  given,  we  may  com- 
pute the  tangent  of  their  sum  or  difference.  To  find  the  cotangent 
of  the  sum  or  difference  when  the  cotangents  of  the  angles  are  given, 
divide  (120)  by  (119), 

,          N      cot  y  cot.-eq=  1  /io-\ 

cot(a?±:y)=-  (12^) 

cot  y  ±  cot  x 

63.  Dividing  (115)  by  (116),  and  (117)  by  (118),  (or  from  the 
equations  of  Art.  61),  we  have 

sin  (x  -f-  y)  _  tan  x  +  tan  y  _  cot  y  -\-  cot  x 

.     ---     —  •  --  —  —  !  —  —  -  • 

sin  (x  —  y)       tan  x  —  tan  y      cot  y  —  cot  x 

cos  (x  -{-  y)  _  1  —  tan  x  tan  y  _  cot  x  cot  y  —  1        /1  0_, 

—  •         (  i  Z  i  J 
cos  (x  —  y)       1-4-  tan  x  tan  y       cot  x  cot  y  -f  1 

64.  Formulae  for  secants  are  obtained  from  those  for  cosines  by  means  of  (2)  ;  thus 
we  find 


sec  (x  ±  y) 


cos  x  cos  y  =F  sin  x  sin  y 
and  multiplying  numerator  and  denominator  by  sec  x  sec  y, 

/  s  sec  x  secy 

sec(x±y)  =  —  — a — 

1  =F  tan  x  tan  y 


34  PLANE  TRIGONOMETRY. 

Also  since 

1  1         cos  v  =fc  COST 

sec  x  ±  sec  y  —  -  ±  -  =  -  —  a  — 

cos  x      cos  y        cos  x  cos  y 


we  find  by  (107)  and  (108) 


2  OOB  *('  +  y)  °"  *('-») 


sec  x  +  sec  y  =  -  (129) 

cos  x  cos  y 

sec  x  -  sec  y  =  2  sin  *  (z  +  ^  sin  *-lx—  «>  (130) 

cos  x  cos  y 

In  the  same  manner  from  (105)  and  (106) 

2  sin  J  (x  -4-  1/)  cos  i  (x  —  V)  /ioi\ 

cosec  x  -}-  cosec  y  =  —  —  "-5  —  r—  *"-  —  1LJ  -  *'  (131) 

sin  x  sin  y 

cosec  x  -  cosec  y  =  -  *™±-(*±-ti*™A(*=3Ll  (132) 

sin  x  sin  y 

These  formulae,  although  generally  omitted  in  treatises  on  trigonometry,  will   be 
found  useful  in  a  subsequent  part  of  this  work. 

65.  The  product  of  (36)  and  (37),  and  of  (38)  and  (39),  are 

sin  (x  -\-  y)  sin  (x  —  y)  =  sin2  x  cos2  y  —  cos2  x  sin2  y 
cos  (x  -f-  y)  cos  (x  —  y)  —  cos2  x  cos2  y  —  sin2  x  sin2  y 

By  (13)  we  have  cos2  x  —  1  —  sin2  x  and  cos2  y  =  1  —  sin2  y,  which 
substituted  in  the  preceding  equations,  give 

sin  (x  -f-  y]  sin  (#  —  y]  =  sin2  #  —  sin2  y  =  cos2  y  —  cos2  x         (133) 
cos  (x  -f-  i/)  cos  (x  —  y)  =  cos2  #  —  sin2  y  —  cos2  2/  —  sin2  #        (1  34) 

66.  In  (36),  (38)  and  (123),  let  y  =  x,  we  find 

sin  2  x  —  2  sin  x  cos  a;  (1  35) 

cos  2  #  =  cos2  z  —  sin2  x  (1  36) 

2  tan  x 

tan  2  a?  =  (130 

1  —  tan2  x 

by  which  the  functions  of  the  double  angle  may  be  found  from  those 
of  the  simple  angle. 

67.  To  find  the  functions  of  the  half  angle  from   those  of  the 
whole  angle,  we  have,  from  (13)  and  (136), 

cos2  or  +  sin2  x  =  1 
cos2  x  —  sin2  x  —  cos  2  x 

the  sum  and  difference  of  which  are 

2  cos2  x  =  1  -f  cos  2  x 
2  sin2  x  =  \  —  cos  2  x 


GENERAL  FORMULA.  35 

As  these  express  the  relations  of  an  angle  2  x  and  its  half  x,  their 
meaning  will  not  be  changed  by  writing  x  and  J  x  instead  of  2  #  and 
x;  whence 

2  cos2  %  x  =  1+  cos  x  (138) 

2  sin*  $a;  =  l—  cosx  (139) 

the  quotient  of  which  is 

1~cosa; 


1  +  COS  X 
68.  The  following  may  be  proposed  as  exercises. 


(140) 


cinx=     2  tan  $  x 


1  -j-  tan2  \x       cot  £  x  -f  tan  $  a; 

tanz  =     2tan*f     =-          2        ,  (142) 

1  —  tan*  £  x       cot  J  x  —  tan  j  a; 


tan2  Jx  +  2cotx  tan  £x  —  1  =  0 
tanz  J  x  —  2  cosec  x  tan  \  x  +  1  =  0 

1  —  tan*  J  x 
cos  x  =  : 

1  +  tan9  \  x 

tan  \  x  =  cosec  r  —  cot  x  =  —        •  —  • 
smx 

cot  \  x  =  cosec  x  +  cot  x  =  1  +CO8?: 


sn  x      cos  z 


69.  Several  useful  formula  result  from  the  preceding,  by  intro- 
ducing 45°  or  30°.  If  z  =  45°  in  (36),  (37),  (38),  and  (39),  we 
have,  by  (33), 

sin  (45°  ±y)  =  cos  (45°  +  y)  =  cos  V  ±sm  V  (149) 

I/2 
whence 


tan  (45°  ±y)  =  cot  (45°  +  y)  -  (150) 

cos  y  =F  sin  y 

in  which  either  the  tipper  signs  must  be  taken  throughout,  or  the 
lower  signs  throughout. 

If  we  divide  the  numerator  and  denominator  of  (150)  by  cosy  or 
sin  y, 

1  ±  tan  y        cot  y  ±  1 

tan  (45°  ±  y)  =  -         —£= *_: 

1  T  tan  y        cot  y  +  1 


36  PLANE  TRIGONOMETRY. 

From  this,  by  (57), 

tan(y-45°)=  ta"y~  (152) 

tany+1 

70.  Again,  let  z  =  90°  dry  in  (138),  (139),  (140)  and  (146), 


sin(45°±Jy)=coS(4o° 

=FhO  =  Yv  —  2     ) 

(153) 

tan  (45°  zb  \  y)  =  -* 

/  /I  i  sin^\ 
/  '  1  ^f  sin  y/ 

(154) 

tan  (45°  ±  %  y)  =  - 

From  the  last  we  find 
tan  (45°  +  \  y)  +  tan  (45° 

tan  (45°  -f  $  y)  —  tan  (45° 

cosy          l^siny 

A   1M   9    cop  »* 

(155) 

(156) 
(157) 

5  y  i  —          —  *  sec  y 

cosy 

i\        2  sin  y      n  . 
—  J  y)  =  *  =  2  tan  y 
cosy 

the  quotient  of  which  is 

tan  (45°  +  j  y)  -  tan  (45°  -  $  y)        .  , 

° 


71.  In  (101),  (102),  (103)  and  (104),  let  x  =  30°;   then  by  (27) 
and  (28) 

sin  (30°  +  y)  +  sin  (30°  —  y)  =  cos  y  (1  59) 

sin  (30°  +  y)  —  sin  (30°  —  y)  =  sin  y  y  3  (160) 

cos  (30°  +  y)  +  cos  (30°  —  y)  =  cos  T/  y  3  (161) 

cos(30°  +  t/)  —  cos(30°  —  y)=   -sin?/  (162) 

and  in  a  similar  manner  we  may  introduce  60°;  but  it  is  unneces- 
sary to  extend  these  substitutions,  as  they  involve  no  difficulty,  and 
can  be  made  as  occasion  demands. 

FORMULAE  FOR  MULTIPLE  ANGLES. 
72.  From  (101)  and  (102)  we  have 

sin  (y  -j-  x)  —  2  sin  y  cos  z  —  sin  (y  —  z) 
sin  (y  -j-  x)  =  2  cos  y  sin  z  -f-  sin  (y  —  z) 
in  which  let  y  =  (m  —  1)  z  ;  then 

sin  m.e  =  2  sin  (m  —  1)  z  cos  z  —  sin  (m  —  2)x  (163) 

sin  mz  =  2  cos  (m  —  1)  z  sin  z  -f-  sin  (m  —  2)  z  (164) 

which  arc  the  general  formula?  for  computing  the  sine  of  any  multiple  mz,  from  the 
lower  multiples  (m  —  1)  z  and  (m  —  2)  z,  and  the  simple  angle  z. 


FORMULAE  FOR  MULTIPLE  ANGLES.  37 

If  we  make  m  successively  1,  2,  3,  4,  etc.,  these  formulae  give 
sin     x  =     sin  x  sin  x 

sin  2  x  =  2  sin  x     cos  x  =2  cos  x     sin  x 

sin  3  x  =  2  sin  2  x  cos  x  —  sin  x     =2  cos  2  x  sin  x  -(-  sin  x 
sin  4  x  =  2  sin  3  x  cos  x  —  sin  2  x  =  2  cos  3  x  sin  x  -f-  sin  2  z 
etc.  etc. 

73.  From  (103)  and  (104) 

cos  (y  -\-  x)  =       2  cos  y  cos  x  —  cos  (y  —  x) 
cos  (y  -\-  *)  —  —  2  sin  y  sin  x  -(-  cos  (y  —  x) 
which,  if  we  put  y=(m  —  •  1)  x,  become 

cos  mx  =       2  cos  (m  —  1  )  x  cos  x  —  cos  (m  —  2)  x  (165) 

cos  mx  —  —  2  sin  (m  —  1)  x  sin  x  -)-  cos  (m  —  2)  x  (166) 

If  wi  is  taken  successively  equal  to  1,  2,  3,  4,  etc. 

cos     x  =  cos  x  cos  x 

cos  2  x  =  2  cos  x     cos  a;  —  1  =  —  2  sin  x     sin  x  +  1 

cos  3  x  —  2  cos  2  x  cos  x  —  cos  x     =  —  2  sin  2  x  sin  x  -f-  cos  x 
cos  4  x  =  2  cos  3  x  cos  x  —  cos  2  x  =  —  2  sin  3  x  sin  x  -\-  cos  2  x 
etc.  etc. 

74,  In  (123)  let  y  —  (m  —  1)  x;  then 

tan  wix  =    ta"(m~  l)x  +  tanz 
1  —  tan  (m  —  1)  x  tan  x 
whence 

tan     x  =  tanx  tan  3  x  =  -*an  2  x  +  tanx- 

1  —  tan  2  x  tan  x 

tan2z  =     2tanx-  tan4x= 


1  —  tan1  x  1  —  tan  3  x  tan  x 

etc. 

75.  If  in  the  expression  for  sin  3  x,  Art.  72,  we  substitute  the  value  of  sin  2  x, 
we  find 

sin  3  x  =  4  sin  x  cos*  x  —  sin  x, 

by  which  we  find  the  sine  of  the  multiple  directly  from  the  functions  of  the  simple 
angle.  If  this  be  substituted  in  the  expression  for  sin  4x,  the  latter  will  also  be 
expressed  in  terms  of  the  simple  angle.  By  these  successive  substitutions  we  easily 
obtain  the  following  tables  : 

sin     x  =     sin  x 
sin  2  x  =  2  sin  x  cos  x 
sin  3  x  =  4  sin  x  cos*  x  -  -  sin  x 
sin  4  x  =  8  sin  x  cos*  x  —  4  sin  x  cos  z 
etc. 

76.  cos     z  =     cos  x 

cos  2  x  =  2  cos*  x  —  1 
cos  3  x  =  4  cos5  x  —  3  cos  x 
cos  4  x  =  8  cos*  x  —  8  cos1  z  -j-  1 
etc. 

D 


38  PLANE  TRIGONOMETRY. 

77.  If  in  these  equations  we  substitute  for  cos1  x  =  1  —  sin*  x  they  become 

sin     x  =     sin  z 
sin  2  x  =  2  sin  a:  j/  ( 1  —  sin1  z) 
sin  3  x  =  3  sin  z  —  4  sin3  z 
sin  4  x  =  (4  sin  x  —  8  sin5  x)  j/  (1  —  sin1 2) 
etc. 

78.  cos    x  =  i/  (1  —  sin2  x) 
cos  2  x  =  1  —  2  sin2  x 

cos  3  x  =  (1  —  4  sin2  x)  -j/  (1  —  sin1  x) 
cos  4  x  =  1  —  8  sin2  x  -)-  8  sin*  x 
etc. 

From  the  preceding  tables  it  appears  that  the  cosine  of  the  multiple  angle  may 
always  be  expressed  rationally  in  terms  of  the  cosine  of  the  simple  angle ;  but  that 
the  sine  of  only  thfe  odd  multiples  and  the  cosine  of  only  the  even  multiples  can  be 
expressed  rationally  in  terms  of  the  sine  of  the  simple  angle. 

79.  By  successive  substitutions  we  find  from  the  formulae  of  Art.  74. 

tan     x  =  tan  x 

2  tan  x 


tan  2  x  = 
tan  3  x  = 

tan  4  x  = 


1  —  tan2  x 

3  tan  x  —  tan8  x 
1  —  3  tan1  x 

4  tan  x  —  4  tan8  x 
1  —  6  tan2  x  -j-  tan*  x 

etc. 


80.  The  preceding  results  are  but  particular  applications  of  general  formulae  to  be 
given  hereafter  (Chapter  XV.).  They  are  introduced  here  for  the  convenience  of 
reference  in  elementary  applications.  The  powers  of  the  sine  or  cosine  of  the  simple 
angle  may  also  be  expressed  in  the  multiples  of  the  angle:  but  they  are  most  readily 
obtained  from  the  general  formulae  of  Chapter  XV. 


RELATIONS  OF  THREE  ANGLES. 

81.  Let  x,  y,  and  z  be  any  three  angles  ;  we  have,  by  (36)  and  (38), 
sin  (x  -}-  y  -\-  z)  =  sin  (x  -f-  y)  cos  z  -f-  cos  (x  +  y)  sin  z 
=  sin  x  cos  y  cos  z  -f-  cos  x  sin  y  cos  z 
-\-  cos  x  cos  y   sin  z  —  sin  x  sin  y  sin  z  (168) 

cos  (z  -|-  y  +  z)  =  cos  (x  +  y)  cos  z  —  sin  (x  +  y)  sin  z 
=  cos  x  cos  y  cos  z  —  sin  z  sin  y  cos  z 
—  sin  x  cosy  sin  z  —  cos  x  sin  y  sin  z  (1G9) 

and  in  the  same  way  we  may  develop  the  sines  and  cosines  of  x  +  y  —  *,  x  —  y  +  z, 
etc.  ;  but  we  may  find  these  directly  from  (168)  and  (169)  by  changing  the  sign  of  z, 
y,  etc.,  and  observing  (56). 


RELATIONS  OF  THREE  ANGLES.  39 

The  quotient  of  (168)  divided  by  (169)  gives,  after  dividing  the  numerator  and 
denominator  by  cos  x  cos  y  cos  z, 

,  >          tan  z  +  tan  v  +  tan  z  — tanz  tan  y  tanz  /tnn\ 

tan  (x  +  V  +  z)  =  ~  — ^-^  ( 1 70) 

1  —  tan  x  tan  y  —  tan  z  tan  z  —  tan  y  tan  z 

82.  Let  x,  y  and  z  be  any  three  angles,  and  from  the  equations 

sin  (x  —  z)  =  sin  z  cos  z  —  cos  z  sin  z 
sin  (y  —  z)  =  sin  y  cos  z  —  cos  y  sin  z 

let  cos  z  be  eliminated ;  we  find 

sin  y  sin  (z  —  z)  —  sin  x  sin  (y  —  z)  =  sin  z  (sin  z  cos  y  —  cos  z  sin  y) 

=  sin  z  sin  (z  —  y) 

If  sin  z  is  eliminated,  we  find 

cos  y  sin  (z  —  z)  —  cos  z  sin  (y  —  z)  =  cos  z  sin  (z  —  y) 
These  equations  may  be  more  elegantly  expressed,  as  follows: 

sin  z  sin  (y  —  z)  -f  sin  V  sin  (z  —  x]  +  sin  z  sin  (x  —  y)  =  0  (171 ) 

cosz  sin  (y  —  z)  -f-  cosy  sin  (z  —  x)  -f-  cosz  sin  (z  —  y)  =  0  (172) 

A  number  of  similar  relations  may  be  deduced  from  these  by  substituting  90°  ±:  x, 
etc.,  for  z,  etc. 

83.  Let 

v  =  J  (x  +  V  +  «) 

we  have  by  (104) 

2  sin  v  sin  (v  —  z)  =  cos  z  —  cos  (2  v  —  z)  =  cos  z  —  cos  (y  +  z) 
2  siu  (u  — •  y)  sin  (v  —  z)  =  cos  (y  —  z)  —  cos  (2  v  —  y  —  z) 

=  cos  (y  —  z)  —  cos  z 
the  product  of  which  is 

4  sin  t>  sin  (v  —  z)  sin  (v  —  y)  sin  (v  —  z)  =  cos  z  [cos  (y  —  z)  +  cos  (y  +  z)] 

—  cos2  z  —  cos  (y  -f-  2)  cos  (y  —  z) 
Reducing  the  second  member  by  (103)  and  (134) ; 

4  sin  v  sin  (v  —  z)  sin  (v  —  y)  sin  (v  —  z)  =  2  cos  z  cos  y  cos  z  —  cos1  z 

—  cos'  y  —  cos1  z  +  1  (173) 

In  the  same  manner  we  find 

4  cos  v  cos  (v  —  z)  cos  (v  —  y)  cos  (v  —  z)  —  2  cos  z  cos  y  cos  z  -{-  cos3  z 

+  cos2  y  +  cos1  z  —  1  (174) 

84.  The  following  may  be  proposed  as  exercises. 

sin  z  +  sin  y  +  sin  z  —  sin  (z  -f  y  +  z)  =  4  sin  $(*  +  !/)  sin  J  (*  +  2)  sin  J  (y  -f-  z)  (175) 
cos  z  +  cos  y  -|-  cos  z  -f  cos  (z  +  y  +  z)  =  4  cos  £  (z  +  y)  cos  £  (z  -f  z)  cos  }  (y  +  z)   (176) 

tan  z  +  tan  y  -f  tan  z  —  tan  z  tan  y  tan  z  ^  Rin  1?-+J-+-?)  (177) 

cos  z  cos  y  cos  z 

cot  z  +  cot  y  +  cot  z  —  cot  z  cot  y  cot  z  =  —  eos  (z  +  y  +  ?)      (178) 

sin  z  sin  y  sin  z 


40  PLANE  TRIGONOMETRY. 

4  [sin  (x  -f  y  +  z)  +  2  sin  x  sin  y  sin  z]*  =  4  [sin  (x  +  y)  cos  z  -f  cos  (x  —  y)  sin  z]s ' 

=  [1  —  cos  (2  x  +  2  y)]   (1  +  cos  2  z)  +  [1  +  cos  (2  z  —  2  y)]  (1  —  cos  2  z) 

f  (179) 
+  2  (sin  2  x  +  sin  2  y)  sin  2  z 

=  2  (1  +  sin  2  a;  sin  2  y  -f  sin  2  z  sin  2  z  +  sin  2  y  sin  2  z  —  cos  2  z  cos  2  y  cos  2  z)  . 

85.  Let  the  sum  of  three  angles  x,  y  and  z  be  w,  or  a  multiple  of  »r,  that  is,  an  even 

multiple  of  -  a  condition  which  is  expressed  by  the  equation 

Z 

z  +  y  +  z  =  2n.^  (180) 

2 

then,  tan  (x  +  y  +  z)  =  0,  and  the  first  member  of  (170)  being  thus  reduced  to  zero, 
the  numerator  of  the  second  number  must  be  zero,  or 

tan  x  -f-  tan  y  -f-  tan  z  =  tan  x  tan  y  tan  z  (181) 

an  equation,  it  must  be  remembered,  that  is  true  only  under  the  condition  (180). 
Since  x,  y  and  z  may  be  selected  in  an  infinite  variety  of  ways  so  as  to  satisfy  (180) 
it  follows  from  (181)  that  there  is  an  infinite  number  of  solutions  of  the  problem, 
"  to  find  three  numbers  whose  sum  is  equal  to  their  product." 

Let  the  sum  of  three  angles  x,  y  and  z  be  —  or  an  odd  multiple  of  -  ;   that  is,  let 

•-  — 

x  +  y  +  z  =  (2n  +  l)^  (182) 

then,  tan  (x  -j-  y  -}-  z)  =  oc,  and  the  denominator  of  (170)  must  be  zero,  or 

tan  x  tan  y  +  tan  x  tan  2  -f-  tan  y  tan  z  =  1 
which,  divided  by  tan  x  tan  y  tan  z,  gives 

cot  x  +  cot  y  +  cot  z  =  cot  x  cot  y  cot  z  (183) 

a  relation  that  holds  only  under  the  condition  (182). 

86.  Let 

x-fy  +  z  =  nTT  =  2n-  (184) 

2 

We  have  by  (93)  and  (91) 

cos  (x  -f-  y  —  z)  =  cos  (ra  TT  —  2  z)  =  ( —  1)"  cos  2  z 
cos  (x  —  y  +  z)  =  cos  (n  T  —  2  y)  =  ( —  1)"  cos  2  y 
cos  (y  +  z  —  x)  =  cos  (n  TT  —  2  x)  =  ( —  l)n  cos  2  x 
cos(y  +  z  +  :r)  =  cosn7r       •        =( — 1)B 
the  sums  of  the  first  two  and  of  the  second  two  are  by  (103) 

2  cos  x  cos  (y  —  z)  =  (—  1)B  (cos  2  z  +  cos  2  y) 
2  cos  x  cos  (y  +  2)  =  ( —  l)n  (cos  2x  -f  1) 
and  the  sum  and  difference  of  these  equations  are 

4  cos  x  cos  y  cos  z  =  ( —  1 )"  (cos  2  z  +  cos  2  y  -f  cos  2  x  +  1 ) 
4  cos  x  sin  y  sin  z  =  ( —  1)"  (cos  2z-fcos2y  —  cos2x— 1) 

±4  cos  x  cosy  cos  2=       cos2x4-cos2y-|-cos2z  +  l  (185) 

±  4  cos  x  sin  y  sin  z  =  —  cos  2  x  -\-  cos  2  y  +  cos  2  z  —  1  (1 86) 

the  upper  sign  being  taken  when  n  in  (184)  is  even,  the  lower  when  n  is  odd. 


INVERSE  TRIGONOMETRIC  FUNCTIONS.  41 

In  the  same  manner  we  obtain 

=f  4  sin  x  sin  y  sin  z  —       sin  2  x  +  sin  2y  -(-  sin  2z  (187) 

=F  4  sin  x  cos  y  cos  z  =  —  sin  2  z  -(-  sin  2  y  -f  sin  2  2  (188) 

the  signs  being  taken  as  above. 

Again,  let 

*  +  y  +  *  =  (2n  +  l)^  (189) 

we  shall  6nd  by  the  same  process 

±  4  sin  z  sin  y  sin  z  =       cos  2  z  -(-  cos  2  y  -f  cos  2  z  —  1  (190) 

zb  4  sin  x  cos  y  cos  z  =  —  cos  2  x  -+-  cos  2  y  -f-  cos  2  2  -f  1  (191) 

±  4  cosz  cosy  cosz  —       sin  2x  -\-  sin  2y  -f  sin  2z  (192) 

±  4  cos  x  sin  y  sin  z  =  —  sin  2  z  -f  sin  2  y  -)-  sin  2  z  (193) 

+  or  —  according  as  n  in  (189)  is  even  or  odd. 

INVERSE  TRIGONOMETRIC  FUNCTIONS. 

87.  If 

?/  =  sin  x 

y  is  an  explicit  function  of  x,  and,  since  x  and  y  are  mutually  de- 
pendent, x  is  an  implicit  function  of  y  ;  but  to  express  x  in  the 
form  of  an  explicit  function  of  y,  we  write  * 


which  is  read,  t(  x  equal  to  the  angle  (or  arc)  whose  sine  is  y"  and  x 
is  called  the  inverse  function  of  y,  or  of  sine  x. 

In  like  manner  tan  ~l  y  is  "  the  angle  or  arc  whose  tangent  is 

y,"  etc. 

88.  Many  of  the  formulae  already  given  may  be  conveniently  expressed  with  the 
aid  of  this  notation.    Thus,  by  (16), 


+  tanjz) 
or  if  we  put  y  =  tan  x 

tan  -  *  y  =  sec  ~  l  \/  (1  -f  y1) 

*  This  notation  was  suggested  by  the  nse  of  the  negative  exponents  in  algebra.  If 
we  have  y  —  nz,  we  also  have  x  •=  n  ~  '  y,  where  y  is  a  function  of  z,  and  z  is  the 
corresponding  inverse  function  of  y.  The  latter  equation  might  be  read  "z  is  a  quan- 
tity which  multiplied  by  n  gives  y."  It  may  be  necessary  to  caution  the  beginner 

against  the  error  of  supposing  that  sin-1y  is  equivalent  to  -„•  —  • 

sin  y 

For  a  general  view  of  the  nature  of  inverse  functions,  see  Peirce's  Diff.  Calc.,  Arts. 
13,  ct  seq. 


42  PLANE  TRIGONOMETRY. 

And  in  the  same  way  the  formulae  of  Art.  28  give 

sin  -  *  y  =  cosec  ~  *  -  —  cos  -1i/(l  —  y7)  =  tan  - 1  — 

y  i/(i-y') 

cos ~ly  =     sec  ~ *  —  =  sin  - 


tan~'y=     cot-'-^siu-1 — — * — —  —  cos 

y 


Formulae  (123)  and  (124)  may  be  written 


- 

1  =F  tan  x  tan  y 

or  putting  t  =  tan  z,  i'  —  tan  y, 

tan-1<±tan-1</  =  tan~1   i  —  i'  (194) 

1  =F«' 

Also  the  formulae  of  Arts.  67  and  68  give 

2  «.-  ^/  (L±»)  =,  „„-,  ^ 


sin 


89.  We  may  also  employ  the  notation  sin"1  (cos  x)  or  "the  arc  whose  sine  is  equal 
to  the  cosine  of  r,"  i.  e.  "  the  complement  of  x"  ;  and  sin  (cos"-1  y)  or  "the  sine  of  the 
arc  whose  cosine  is  y,"  etc.  We  shall  have  accordingly 

sin  (sin  -  *  y)  =  y  tan  (tan  ~  '  y)  =  y  etc. 

sin  -  l  (sin  x)  —  x  tan  ~  1  (tan  r)  =  x  etc. 

But  it  must  be  observed  that  since  the  same  sine  or  tangent  corresponds  to  an 
infinite  number  of  angles,  (Art.  53,)  these  last  equations  should  be  written 

sin"1  (sinr)  —  mr  -f   (  —  l)"r,  tan~  '  (tan  x)  —  n  K  -f  x 

which  are  equivalent  to  (95)  and  (97). 


..ad  take 
.uder,  prefixing 

CHAPTER    '16'  =  etc. 

sin  50°  16'  =  etc. 
TKIGONOM 

method,  as  the  subtraction  of  90°  may 
90.  BEFORE  proceedinr 

and  to  other  applicat\3veen  jgQ0  and  270°,  we  deduct  180°  and  take 
make  himself  apifs  of  the  remainder,  prefixing  the  signs  that  belong 
consulting,  tjbdrant,  Art.  41  j  thus 

f°intS  °"'-  sin  220°  26'  =  -  sin  40°  26' 

to  consul 

in  order  cos  220°  26'  =    -  cos  40°  26' 

suppose  tan  220°  26'  ==  +  tan  40°  26'  etc. 

common 

There01*  angles  between  270°  and  360°,  we  may  deduct  270°  and 
Table  <  complemental  functions  of  the  remainder,  prefixing  the  signs 
of  the  iOng  to  the  4th  quadrant,  Art.  43 ;  thus 


sin  331°  27'  =  —  cos  61°  27' 

Th- 

a  cos  331°  27'  =  +  sin  61°  27' 

e  '  tan  331°  27'  =  -  cot  61°  27'  etc. 


selv-)r  we  may  j.ake  £ne  same  functions  of  the  difference  between  the 
t]etele  and  360°,  Art.  44,  observing  the  signs. 

emj'6.  Above  360°  we  deduct  360°,  and  take  the  same  functions  with 
of  qr  signs,  Art.  45;  and  if  the  angle  exceeds  720°,  1080°,  etc.,  we 
de<luct  720°,  1080°,  etc.;  thus 

sar 

cos    840°  45'  =  cos  120°  45'  =     -  sin  30°  45' 
rac  tan  1372°  13'  =  tan  292°  13'  =    -  cot  22°  13' 

of 

TABLE  OF  LOGARITHMIC  SINES,  ETC. 

m 

97.  In  this  table  we  find  the  logarithms  of  the  numbers  in  the 
^ble  of  Natural  Sines  arranged  in  precisely  the  same  manner;  it 
j(ill  therefore  require  but  little  additional  explanation. 
j  As  the  sines  and  cosines  are  all  less  than  unity  (being  by  their 
definitions  proper  fractions),  their  logarithms  properly  have  negative 
ndices;  but  these  are  avoided  in  the  usual  manner  by  increasing  the 


44  PLANE  TRIGONOMETRY. 

second,   called    the    Table    of  Logarithmic   Sines,    etc.,  contains    the 
'thms  of  the  numbers  in  the  first  table.     As  the  greater  part 
of  trigonometry  are  carried  on  by  logarithms, 
far  tji 


tan  ~  '  y  —     cot  ~  *  —  - 

T  TURAL  SINES,  ETC. 

Formulae  (123)  and  (124)  may  be  written 

•  will  be  understood  from  a  sim- 

.          _  ,         _  i      I,  111  Z 

f  =F  tan  x  ?tc.  of  angles  between  zero 
or  putting  t  =  tan  z,  <'  =  tan  y,  ,  'u-   functions  of  angles 

/  ,  //       ™onds,  have  to  be 
tan~l  <±  tan-1  17  =  tan-1   l—  . 

ise  tliat  are  re- 


Also  the  formula;  of  Arts.  67  and  68  give  Mlterpolation 

i  /i  _,A  ,  ,-i   I  ,,\  /  .sines,  etc., 

cos-^2  sin-  V  (V)  =2  C°S-]  V  m  ^  tan"  V  (v   piopor- 


2  tan 


-_ 

1  +  f  l  —  y"  V  1  + 

89.  We  may  also  employ  the  notation  sin"1  (cos  x)  or  "the  arc  whose  sinefi  tables 
to  the  cosine  of  r,"  i.e.  "the  complement  of  x";  and  sin  (cos~*y)  or  "the  siiO",  and 
arc  whose  cosine  is  y,"  etc.  We  shall  have  accordingly 

sin  (sin  -  l  y)  =  y  tan  (tan-1  y)  —  y  etc.  lent  of 

sin  ~  l  (sin  x)  =  x  tun-1  (tan  x)  •=  x  etc.  n   45° 

But  it  must  be  observed  that  since  the  same  sine  or  tangent  corresponds  3Otail- 
infinite  number  of  angles,  (Art.  53,)  these  last  equations  should  be  written  ^tend 

sin-1  (sin  r)  =  n  *•  +  (—  I)"*,  tan~  '  (tan  or)  —  n  *  +  z       45°, 

which  are  equivalent  to  (95)  and  (97). 

'cal 
not 
ler 


<e 
.er 


TABLE  OF  TRIGONOMETRIC  SINES.  45 

remembering  that  in  the  2(1  quadrant  all  the  functions  are  negative 
except  the  sine,  and  its  reciprocal,  the  cosecant. 

Or,  we  may  (Art.  38)  deduct  90°  from  the  given  angle  and  take 
from  the  table  the  complemental  functions  of  the  remainder,  prefixing 
the  signs  as  before ;  thus 

sin  140°  16'  =       cos  50°  16'  —  etc. 
cos  140°  16'  =    -  sin  50°  16'  =  etc. 

which  is  the  better  practical  method,  as  the  subtraction  of  90°  may 
be  performed  mentally. 

94.  For  angles  between  180°  and  270°,  \ve  deduct  180°  and  take 
the  same  functions  of  the  remainder,  prefixing  the  signs  that  belong 
to  the  3d  quadrant,  Art.  41 ;  thus 

sin  220°  26'  =  —  sin  40°  26' 
cos  220°  26'  -  -  cos  40°  26' 
tan  220°  26'  ==  +  tan  40°  26'  etc. 

95.  For  angles  between  270°  and  360°,  we  may  deduct  270°  and 
take  the  complemental  functions  of  the  remainder,  prefixing  the  signs 
that  belong  to  the  4th  quadrant,  Art.  43;  thus 

sin  331°  27'=- -cos  61°  27' 
cos  331°  27'  =  +  sin  61°  27' 
tan  331°  27'  =  -  cot  61°  27'  etc. 

Or  we  may  take  the  same  functions  of  the  difference  between  the 
angle  and  360°,  Art.  44,  observing  the  signs. 

96.  Above  360°  we  deduct  360°,  and  take  the  same  functions  with 
their  signs,  Art.  45;  and  if  the  angle  exceeds  720°,  1080°,  etc.,  we 
deduct  720°,  1080°,  etc.;  thus 

cos    840°  45'  =  cos  120°  45'  =     -  sin  30°  45' 
tan  1372°  13'  =  tan  292°  13'  -    -  cot  22°  13' 

TABLE  OF  LOGARITHMIC  SINES,  ETC. 

97.  In  this  table  we  find  the  logarithms  of  the  numbers  in  the 
Table  of  Natural  Sines  arranged  in  precisely  the  same  manner;  it 
will  therefore  require  but  little  additional  explanation. 

As  the  sines  and  cosines  are  all  less  than  unity  (being  by  their 
definitions  proper  fractions),  their  logarithms  properly  have  negative 
indices;  but  these  are  avoided  in  the  usual  manner  by  increasing  the 


46  PLANE  TRIGONOMETRY. 

index  by  10,  so  that  we  find  the  index  9  instead  of  --  1,  8  instead 
of  —  2,  etc.  The  tangents  under  45°  being  also  less  than  unity,  the 
indices  of  their  logs,  are  also  increased  by  10. 

In  some  tables,  to  preserve  uniformity,  the  indices  of  the  logs,  of 
all  the  functions  are  increased  by  10,  so  that  the  log.  secants  and 
cosecants  are  always  greater  than  10.  In  using  these  tables,  we  have 
the  general  rule  that  for  each  log.  function  added  in  forming  a  sum 
we  must  deduct  10  from  that  sum. 

98.  Since 


we  have  log  sec  A  =  —  log  cos  A 

or  since  the  tabular  log.  cos.  is  increased  by  10, 
log  sec  A  =  10  —  log  cos  A 

that  is,  the  log.  secant  is  the  arithmetical  complement  of  the  tabular  log. 
cosine.     For  a  like  reason  log.  cosec.  is  the  ar.  co.  of  the  log.  sin. ; 
and  log.  cot.  is  the  ar.  co.  of  the  log.  tan. 
Also  since 

sin  A 
tan  A  = 

cos  A 

log  tan  A  =  log  sin  A  —  log  cos  A 

by  which  property,  together  with  the  preceding,  we  may  obtain  by 
subtraction  only,  the  log.  tan.  cot.  sec.  and  cosec.  from  a  table  con- 
taining only  the  log.  sin.  and  cos. 

99.  When  the  natural  sines,  etc.  are  negative,  we  shall  in  this 
work  indicate  it  by  prefixing  the  negative  sign  also  to  their  logar- 
ithms ;  *  thus  we  shall  write 

cos  140°  16'  =    -  0-7690278 
and  log  cos  140°  16'  =    -  9-8859420 

*  Strictly  speaking,  negative  numbers  have  no  logarithms,  but  in  practice,  the  multi- 
plication, division,  etc.  of  numbers  is  performed  without  reference  to  their  signs,  i.  e. 
as  if  they  were  all  positive,  and  the  sign  of  the  result  is  then  deduced  from  the  signs 
of  the  factors  according  to  the  rules  of  algebra.  We  employ  logarithms  simply  to 
effect  the  first  of  these  operations,  i.  e.  the  multiplication,  division,  etc.  of  the  num- 
bers considered  as  positive;  and  to  facilitate  the  second  operation,  or  the  determination 
of  the  sign,  we  prefix  to  the  logs,  the  signs  which  are  prefixed  to  the  numbers  to  which 
they  belong. 


CONSTRUCTION  OF  TRIGONOMETRIC  TABLES.  47 

As  the  logs,  of  all  the  trig,  functions  are  positive  (being  rendered  so 
by  the  addition  of  10  where  necessary),  it  will  easily  be  remembered 
that  the  sign  in  the  latter  case  is  not  that  of  the  logarithm,  but  of  the 
number  to  which  it  belongs. 

ELEMENTARY  METHOD  OP  CONSTRUCTING  THE  TRIGONOMETRIC 

TABLE. 

100.  By  dividing  TT  =  3*141 5926  by  the  number  of  seconds  in 
180°,  we  found  (Art.  9)  the  length  of  the  arc  1",  and  (Art.  54),  the 
sine  of  1",  which  is  sensibly  equal  to  the  arc.  In  the  same  manner 
we  find,  by  dividing  by  10800, 

sin  1'  =  0-0002908882 
and  by  (7) 

cos  \'=y(l  —  sin2 1 ')=-,/  [(1  +  sin  1')  (1  —  sin  1')] 

=  V  (1-000290888 2  X  -9997091118) 
or  performing  the  arithmetical  operations 

cos  V  =  0-9999999577 
Then  by  (101)  and  (103) 

sin  (x  -\-  y)  =  2  sin  x  cos  y  —  sin  (x  —  y) 
cos  (x  -+-  y)  =  2  cos  x  cos  y  —  cos  (x  —  y) 

in  which  we  can  suppose  y  to  be  constantly  equal  to  V  and  x  to  be- 
come successively  1',  2',  3',  etc.  Thus,  first  substituting  y  —  1', 

sin  (x  +  1')  =  2  sin  x  cos  1'  —  sin  (x  —  1') 
cos  (x  H-  1')  =  2  cos  x  cos  1'  —  cos  (x  —  I/) 
then  if  x  —  1',  2',  3',  etc.,  we  find  for  the  sines 

sin  2'  =  2  cos  V  sin  V  —  sin  0'  =  0-0005817764 
sin  3'  =  2  cos  V  sin  2'  —  sin  V  —  0*0008726646 
sin  4'  =  2  cos  1'  sin  3'  —  sin  2'  =  0-0011635526 
sin  5'  =  2  cos  1'  sin  4'  —  sin  3'  =  0*0014544407 

etc.  etc. 

and  for  the  cosines 

cos  2'  —  2  cos  V  cos  V  —  cos  0'  =  0-9999998308 
cos  3'  =  2  cos  V  cos  2'  —  cos  1'  =  0-9999996193 
cos  4'  =  2  cos  V  cos  3'  —  cos  2'  =  0-9999993232 
cos  5'  =  2  cos  V  cos  4'  —  cos  3'  =  0-9999989425 
etc.  etc. 


48  PLANE  TRIGONOMETRY. 

The  whole  difficulty  in  this  operation  consists  in  the  multiplication 
of  each  successive  sine  or  cosine  by  2  cos  V  =  1*9999999154;  but 
this  multiplication  is  much  shortened  by  observing  that 

2  cos  V  =  1-9999999154  =  2  —  -0000000846 
so  that  if  we  put 

m  =  -0000000846 

we  have  2  cos  V  =  2  —  m  and  therefore 

sin  2'  =  2  sin  V  —  sin  0'  —  m  sin  V 
sin  3'  =  2  sin  2'  —  sin  1'  —  m  sin  2' 
sin  4'  —  2  sin  3'  —  sin  2'  —  m  sin  3' 

etc. 

cos  2'  =  2  cos  V  —  cos  0'  —  m  cos  1 ' 
cos  3'  =  2  cos  2'  —  cos  1' —  m  cos  2' 
cos  4'  =  2  cos  3'  —  cos  2'  —  m  cos  3' 

etc. 

which  are  computed  with  great  facility. 

101.  It  is  not  necessary,  however,  to  continue  this  process  beyond 
30°  ;  for  by  (159)  and  (162)  we  have 

sin  (30°  +  y)  =  cos  y  —  sin  (30°  —  y) 
cos  (30°  +  y)  =  cos  (30°  —  y}  —  sin  y 

so  that  the  table  is  continued  above  30°  by  the  simple  subtraction 
of  the  sines  and  cosines  under  30°  previously  found.  Thus,  making 
it  successively  1',  2',  3',  etc. 

sin  30°  V  =  cos  V  —  sin  29°  59' 
sin  30°  2'  =  cos  2'  —  sin  29°  58' 
sin  30°  3'  =  cos  3'  —  sin  29°  57' 

etc. 

cos  30°  V  =  cos  29°  59'  —  sin  1' 
cos  30°  2'  =  cos  29°  58'  -  sin  2' 
cos  30°  3'  =  cos  29°  57'  —  sin  3' 

etc. 

This  last  process  requires  to  be  continued  only  to  45°,  since  the 
sines  and  cosines  of  the  angles  above  45°  will  be  respectively  the 
cosines  and  sines  of  their  complements  below  45°.' 


CONSTRUCTION  OF  TRIGONOMETRIC  TABLES.  49 

102.  The  tangents  and  cotangents  may  be  found  from  the  sines 
and  cosines  by  the  formulae 

sin  x  cos  x 

tan  x  = cot  x  =  —. — 

cos  x  sin  x 

and  the  secants  and  cosecants  by  the  formulae 

1  1 

sec  x  =  -  cosec  x  =  — — 


cos  x  sin  x 

103.  In  so  extended  a  computation  as  the  construction  of  the  en- 
tire table,  it  is  necessary  to  verify  the  accuracy  of  the  work  from 
time  to  time,  by  separate  and  independent  calculations.     By  means 
of  (138)  and  (139)  we  can  find  from  the  cosine  of  an  angle  the  sine 
and  cosine  of  its  half;  hence  from  the  cos.  45°  =  y  |  we  can  find 
sin.  and  cos.  of  22°  30',  and  from  these  the  sin.  and  cos.  of  11°  15' 
by  the  same  formulae  ;  and  from  cos.  30°  =  £  y  3  we  can  find  sin. 
and  cos.  of  15°,  7°  30',  and  3°  45'.    If  these  agree  with  those  found 
by  the  first  process,  the  whole  work  may  be  considered  as  correct. 

104.  There  are  various  other  angles  whose  functions  can  be  expressed  under  finite 
forms  more  or  less  simple,  and  may  therefore  be  employed  for  the  purpose  of  verifi- 
cation. 

Let  x  =  18°  ;  then  3  x  +  2  x  =  90°  and  cos  3  x  =  sin  2  x,  whence,  by  Art.  76, 

4  coss  x  —  3  cos  x  =  cos  3  x  =  2  sin  x  cos  x 
4  cos*  x  —  3          =2  sin  x 
4  (1  —sin8  a;)  —  3  =  2  sin  x 
sin*  x  -\-  J  sin  x     =  \ 

which  equation  of  the  2d  degree  being  resolved,  gives  sin  z  = 
sin  18°  =  cos  72° 


4 
whence  cos  18°  =  sin  72°  =  V 


From  these  by  (138)  and  (139),  we  find  the  sine  and  cosine  of  9°  and  36°  ;  then 
by  (37)  and  (39)  those  of  36°  —  30°  =  6°,  whence  those  of  3°;  after  which  it  will 
be  easy  to  form  a  table  of  the  exact  values  of  the  sines  and  cosines  for  every  3°  of 
the  quadrant.*  These  expressions,  however,  are  not  of  much  use,  directly,  in  the  con- 
struction of  tables,  as  we  have  much  better  methods  ;  but  they  lead  to  a  formula  of 
verification  which  is  of  some  importance.  We  find 

cos  36°  = 


A  table  of  this  kind  is  given  by  Cagnoli  in  his  Trigonometric. 
£ 


50  PLANE  TRIGONOMETRY. 

And  by  (102) 

fr  •« 

sin  (72°  +  y)  —  sin  (72°  —  y)  =  2  cos  72°  sin  y  =  *£2     -  sin  y 

2 

sin  (36°  +  y)  —  sin  (36°  —  y)  =  2  cos  36°  sin  y  =  V  5  +  *  sin  y 

the  difference  of  these  equations  gives 

sin  (36°  +  y)  —  sin  (36°  —  y)  =  sin  (72°  +  y)  —  sin  (72°  —  y)  +  sin  y 

which  is  Euler* s  formula  of  verification.  By  giving  y  any  value  at  pleasure,  the  cor- 
rectness of  five  sines  of  the  tables  is  examined.  By  substituting  90°  —  y  for  y  in  this 
formula  it  is  easily  reduced  to  the  following 

sin  (90°  —  y)  +  sin  (18°  +  y)  +  sin  (18°  —  y)  =  sin  (54°  +  y)  +  sin  (54°  —  y) 

which  is  known  as  Legendre's  formula,  though  not  essentially  different  from  Euler's. 

105.  The  method  that  has  here  been  given  for  computing  the  trigonometric  table, 
though  simple  in  principle  is  nevertheless  sufficiently  operose.  The  method  by  infi- 
nite series,  to  be  given  hereafter,  will  be  found  to  be  much  more  rapid  and  simple  in 
practice. 


CHAPTER    VI. 

SOLUTION  OF  PLANE  RIGHT  TRIANGLES. 

106.  IN  order  to  solve  a  plane  right  triangle,  two  parts  in  addition 
to    the    right  angle  must  be  given,  one  of  which  must  be  a  side. 
The  solution  is  effected  directly  by  means  of  our  FIG.  15. 
definitions  of  sine,  etc.,  which  are  expressed  by 

the  equations  (1).     As  three  of  the  six  functions 

are  only  the  reciprocals  of  the  other  three,  we 

shall  base  the  solutions  upon  the  following  three;  * 

(Fig.  15): 

a  A       b  .       a 

sin  A  =  -  cos  A  =  tan  A  =  7 

c  c  b 

Since  each  of  these  equations  expresses  a  relation  between  three 
parts — an  angle  and  two  sides — it  follows  that  in  order  to  apply 
them,  or  in  order  to  solve  the  triangle  trigonornetrically,  there  must 
be  given  two  of  these  parts ;  and  that  of  the  three  parts  considered, 
one  must  be  an  angle  while  the  other  two  are  sides.  Thus,  if  an 
angle  and  side  are  given,  the  third  part  sought  must  be  a  side ;  but 
if  two  sides  are  given,  the  third  part  sought  must  be  an  angle. 

In  every  instance  the  choice  of  the  proper  equation  will  be  deter- 
mined by  the  precept, — employ  that  trigonometric  function  of  the  angle 
which  is  equal  to  the  ratio  of  the  two  sides  considered. 

107.  CARE  I.     Given  the  hypotenuse  and  one  angle,  or  c  and  A. 

To  find  a.  We  consider  a,  c  and  A ;  and  since  the  ratio  of  a  and 
c  is  given  by  the  sine,  we  have 

sin  A  =  -     whence     a  =  c  sin  A 
c 

To  find  b.  Considering  b,  c  and  A,  we  have  the  ratio  of  b  and  c 
expressed  by  the  cosine,  or 

cos  A  =  ~     whence     b  =  c  cos  A 
c 

To  find  B,     We  have  B  =  90°  —  A. 


52  PLANE  TRIGONOMETRY. 

The  required  quantities  a  and  b  being  equal  to  the  product  of  two 
factors,  the  computation  is  conveniently  performed  by  the  addition 
of  the  logarithms  of  these  factors. 

EXAMPLES. 

1.  Given  c  =  672-3412,  A  =  35°  16'  25";  to  find  the  other  parts. 

By  (195)  By  (196) 

c  =  672-341 2  log  2-8275897  log  2-8275897 

A  =  35°  16'  25"  log  sin  9-7615382  log  cos  9-9119049 

a  =  388-2647  log*  2-5891279       b  =  548-9018  log*  2-7394946 

Ana.  a  =  388-2647 
b  =  548-9018 
B  =  54°  43'  35" 

2.  Given  c  =  42567-2,  B  —  87°  49'  10"  ;  find  the  other  parts. 

Am.  o=  1619-626 
b  =  42536-37 
A  =     2°  10'  50" 

108.  CASE  II.     Given  the  hypotenuse  and  one  side,  or  c  and  b. 
To  solve  this  case  trigonometrically,  we  must  first  find  an  angle. 
To  find  A.     We  have 

cos^=-  (197) 

c 

To  find  a.     We  have,  by  the  preceding  case, 

sin  A  =  -  «  =  csin^4  (198) 

c 

But  a  may  be  found  by  geometry  from  the  equation 

a2  -f  b2  =  c2     whence     a2  =  c2  —  b2 
a  =  y  (c2  -  62)  =  y  [(c  +  6)  (c  -  &)]  (199) 

EXAMPLES. 
1.  Given  c  =  672-3412,  b  =  548-9018  ;  find  A  and  a. 

By  (197)  By  (198) 

b  ==  548-9018          log  2-7394946 
c  —  672-3412  log  2-8275897  log  2-8275897 

A  =  35°  1 6'  25"  log  cos  9-9119049  log  si n  9-7615382 

a  =±  388-2647  log  2-5891279 

*  Ten  is  rejected  from  each  of  these  indices  because  the  logarithms  of  the  sine  and 
cosine  in  the  table  are  ten  too  great.  Art.  97. 


SOLUTION  OF  PLANE  EIGHT  TRIANGLES.  53 

By  (199) 
c=    672-3412 
b=    548-9018 

c  +  b  =  1221-2430     log  3-0868021 

c  —  b=    123-4394     log  2-0914538 

2)5vTT826_5_9 

a  =  388-2647  log  2-5891279 

Ana.  A  =  35°  16'  25" 
B  =  54°  43'  35" 
a  =  388-2647 
2.  Given  c  =  -092357,  6  —  -058018  ;  find  a. 

4««.  a  =  -071869 

109.  CASE  III.   Given  an  angle  and  its  adjacent  side,  or  A  and  b. 
To  find  a.     We  have 

tan  A  =  -       whence       a  =  b  tan  A.  (200) 

b 

To  find  c.     We  have 

cos  A  =     -  whence,  by  (2),      c  —  -      -  —  b  sec  A  (201) 

c  cos  A 

or  directly  from  the  secant 

sec  A  =  --       whence       c  =  b  sec  A 
b 

EXAMPLES. 

1.  Given  A  =  88°  59',  6  =  2-234875  ;  find  the  other  parts. 

An*.  a=  125-9365 
c  =  125-9563 
£  =  l°  V 

2.  Given  B  =  60°,  a  =  10  ;  find  c.     (See  Art.  29.) 

Ans.  G  =  20. 

110.  CASE  TV.     Given  an  angle  atid  its  opposite  side,  or  A  and  a. 
To  find  c.     We  have 

sin  A  =  -  c—-          -a  cosec  A         (202) 

c  sin  .1 


To  find  b.     We  have 


l>=  =acotA  (203) 

6  tan  A 


E  2 


54  PLANE  TRIGONOMETRY. 

EXAMPLES. 

1.  Given  A  =  25°  18'  48",  a  =  -085623  ;  find  b. 

Ans.  6  =  -1810278 

2.  Given  B  =  39°  17'  5",  6  =  -01  ;  find  c. 

^4«s.  c  =  -0157934 

111.  CASE  V.     Given  the  two  sides,  or  a  and  6. 
To  find  A  and  B.     We  have 


»  (204) 

6 


To  find  c.     We  have 


sin  A  =  -  c  =  -      -  =  a  cosec  A  (205) 

c  sin  A 

We  may  also  find  c  directly  by  geometry,  from 

c2  =  a2  -f  b'z       whence       c  =  y  (a2  +  62) 
but  this  is  not  readily  computed  by  logarithms. 

EXAMPLES. 

1.  Given  a  =  30,  6  =  40  ;  find  c.  Ans.  c  =  50 

2.  Given  a  =  8*678912,  6  =  2-463878  ;  find  A  and  c. 

^ns.  ^4  =  74°  9'  4"  -1 
c  =  9-021875 

ADDITIONAL  FORMULA  FOR  RIGHT  TRIANGLES. 

112.  By  inspecting  the  tables  it  will  be  seen  that  when  the  angles  are  very  small, 
the  cosines  differ  very  little  from  each  other ;  consequently  a  small  angle  cannot  be 
found  with  very  great  accuracy  from  its  cosine.     For  a  similar  reason  an  angle  that  is 
nearly  90°  cannot  be  accurately  computed  from  its  sine.     It  is  therefore  desirable,  when 
a  rf -quired  angle  is  small,  to  find  it  by  its  sine,  and  when  near  90°  by  its  cosine,  or  in 
either  case  by  its  tangent  or  cotangent ;  and  for  this  purpose  special  formula  are  some- 
times necessary.     We  shall  deduce  several  such  formulae,  from  which  one  adapted  to  a 
particular  case  may  be  selected. 

113.  From  (197)  we  find,  by  (139) 

1  —  cos  A  =  2  sin2  \  A  =  1  — 

c  c 

(206) 

which  may  be  used  instead  of  (197),  when  A  is  small,  that  is  when  b  is  nearly  equal 
to  c.     It  gives  also 

c  —  b  -—  2  c  sin2  £  A  (207) 

by  which  c  —  b  may  be  accurately  found  when  A  is  small. 


ADDITIONAL  FORMULA  FOR  RIGHT  TRIANGLES.  55 

Also  from  (197),  by  (140) 

1     A  //I  —  COS  A\  lid  -  b\          C  -  b  /nno\ 

tanM  =  \l7-7  --  -.)  =  A/(~TT)  =  -  (208) 

\  \l-\-ctmAJ       \  Vc  +  6  /          a 
114.  From  the  equation 

sin  A  =  - 
c 

we  deduce  by  (153)  and  (154) 

sin  (45°  ±*A)  =  )  (209) 

2  (210) 


tan  (45°  ±$A)=J  M^)  (211) 

\  \c  =F  af 


and  from  tan  .4  =  y  we  find  by  (151), 
6 


tan  (45°  ±  A)  =        ^  (212) 

6  =F  a 


115.  By  (136)  we  have 

cos  2  -4  =  cos*  J.  —  sin*  .4  = 


c1 
which,  since  2  ^  =  A  +  90°  —  B  =.  90°  —  (.6  —  A),  gives 


By  (135) 


cos  (B  —  A)=  sin  2^  =  2  sin  4  cos  A  =  (214) 

c 


and  from  (213)  and  (214) 

tan  (B-A)  =  MLrj*i_VL=i2L  (215) 


by  wliich  5  —  ^4  is  found  with  great  accuracy  when  6  and  a  are  nearly  equal. 
EXAMPLE.    Given  c  =  4602'836,  6  =  4602-21059  to  find  A. 

By  (206). 

c  —  6  =  0  62541  log  9-7961648 

2  c  =  9205-672  log  3-9640555 

2)5-8321093 

\A  =  28'  20//-18  log  sin  \A  =  7'9160547 

A  =  56'  40/A36 

The  ordinary  process  gives  log  cos  A  =  9'9999410,  whence  A  =  56r  40//.  Tliese 
results  are  obtained  by  Stanley's  Tables,  in  which  the  log.  sines,  etc.,  are  given  for 
every  10"  for  the  first  15°.  A  greater  discrepancy  between  the  two  results  would  be 
found  by  tables  in  which  the  functions  were  given  only  for  each  minute. 

A  slight  error  remains  in  the  value  of  £  A  =  28'  20//-18,  on  account  of  the  large 
differences  of  the  log.  sines  in  this  part  of  the  table,  or  rather  on  account  of  the 


56  PLANE  TRIGONOMETRY. 

rapid  change  of  these  differences.  We  avoid  the  use  of  these  large  differences,  and 
gain  somewhat  in  accuracy,  by  employing  the  approximate  value  of  sin.  £  A  given  by 
(98),  whence 


sin  ^A  —  ^Asmlf/t  %A  =  sm  * 

sin  1" 

Thus  we  have  found  above  log  sin  J  A  =  7'9160547 

Art.  54,  log  sin  1"  =  4*6855749 
\  A  =  1700"-12  =  28'  20/A12  log  i  ^4  =  3-2304798 

But  to  obtain  J  A  with  the  utmost  precision,  recourse  must  be  had  to  the  following 
process,  which  is  constantly  employed  in  observatories,  and  wherever  small  angles  are 
to  be  computed  with  extreme  accuracy.  Special  tables  are  prepared  containing  for 
every  minute  from  0°  to  2°  the  logarithms  of 

*™H=h          and 


which  do  not  vary  rapidly,  and  may  therefore  be  taken  with  accuracy  from  the  tables. 
Then  we  have 


tan  x  =  x . 


A 
tan  x 


A  table  of  this  kind  will  be  found  on  page  156  of  Stanley's  Tables,  where  the 
notation  used  is 

q  •=  log  sin  x,  n  =  log  x 

and  therefore  in  the  column  marked  q  —  n  we  find  the  log  51IL?.    Thus  in  the  above 

x 

example  we  have  found  log  sin  J  A  =  q        =  7%91  60547 

and  from  the  table  q  -n  =  4*6855700 

}  A  =  1700"'14  =  28'  20/A14  log  }  A  =       n  =  3'2304847 

which  is  the  true  value  of  J  A  within  0//-01. 
Stanley's  Table  contains  also  the  values  of 


log                 -  q       n 

X 

n  —  lug  i.) 

1°B   '  •      '  —  7  ~l~  n 
sin  x 

i               X 
\n(t   —  n  4-  n 

(n  —  Inrr  rot,  r,. 

the  use  of  which  may  easily  be  inferred  from  the  example  just  given. 


CHAPTER   VII. 

FORMULAE  FOR  THE  SOLUTION  OF  PLANE  OBLIQUE  TRIANGLES. 

116.  As  every  oblique  triangle  may  be  resolved  into  two  right 
triangles  by  a  perpendicular  from  one  of  the  angles  upon  the  opposite 
side,  we  are  enabled  to  deduce  all  the  formulae  for  their  solution  from 
those  of  the  preceding  chapter. 

117.  The  sides  of  a  plane  triangle  are  proportional  to  the  sines  of 
their  opposite  angles. 

Denote  the  angles  of  the  triangle  ABC,  nG'  16>  c 

Fig.  16,  by  A,  B  and  C,  and  the  sides  oppo-  ^^**XT\ 

site  these  angles  respectively  by  a,  6  and  c.         ^^*       M  \ 
From   C  draw  C  P  perp.  to  A  B  and   put      A*  p 

CP  =  p.      Then  in  the  right  triangles    AGP,  BCP,  we  have, 
by  (195) 

p  =  b  sin  A,  p  =  a  sin  B 

whence  b  sin  A  ~  a  sin  B 

which,  converted  into  a  proportion,  gives 

a  :  b  =  sin  A  :  sin  B  (216) 

and  in  the  same  way  we  may  prove  that 

a  :  c  =  sin  A  :  sin  C 
b  :  c  —  sin  B  :  sin  C 

and  these  three  proportions  may  be  written  as  one,  thus  : 

a  :  b  :  c  =  sin  A  :  sin  B  :  sin  C 


or  thus,  --  =  --  =  -^—  (218) 

sin  A       sin  B      sin  C 

When  the  perpendicular  falls  without  the  c 

triangle,  Fig.  17,  the  angle  CB  P  is  the  sup- 
plement of  J5,  but  by  Art.  39,  it  has  the  same 
sine,  so  that  the  triangle  C  B  P  gives 

p  —  a  sin  CB  P=  a  sin  B 

8  57 


58  PLANE  TKIGONOMETKY. 

the  same  as  was  found  from  Fig.  16.  The  proposition  is  therefore 
general  in  its  application.* 

118.  The  sum  of  any  two  sides  of  a  plane  triangle  is  to  their  differ- 
ence as  the  tangent  of  half  the  sum  of  the  opposite  angles  is  to  the 
tangent  of  half  their  difference. 

For,  by  the  preceding  article, 

a  :  b  =  sin  A  :  sin  B 
whence,  by  composition  and  division, 

a  +  b  :  a  —  b  =  sin  A  +  sin  B  :  sin  A  —  sin  B 
But  from  (109)  if  x  =  A,  y  =  B  we  obtain  the  proportion 
sin  A  +  sin  B  :  sin  A  —  sin  .B  =  tan  ^  (A  +  B)  :  tan  ^  (A  —  B) 

which,  compared  with  the  above,  gives 

a  +  b:a  —  b  =  teu$(A  +  B):tan%(A  —  B)  (219) 

This  may  also  be  written 

o+          ten      ^  +  3 


a  -  b      tan  ±(A  -  B) 


(220) 


and  we  may  infer  the  same  relation  between  b,  c,  B,  C  and  a,  c, 
A,  C. 

119.  The  square  of  any  side  of  a  triangle  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides  diminished  by  twice  the  rectangle  of  these 
sides  multiplied  by  the  cosine  of  their  included  angle. 

In  the  triangle  ABC,  Figs.  16  and  17,* 
we  have  either 

BP=c-AP       orBP  =  AP—c 


p 
FlQ  17  but  in  both  cases 

BP*  =  AP2  +  c*—  2cXAP 
Adding  C  P2  to  both  members,  we  find 


But  the  triangle  A  CP  gives  by  (196) 
A  P  =  b  cos  A 


*The  consideration  of  Fig.  17  was  not  strictly  necessary  according  to  the  principle 
stated  in  Art  49.  It  may,  however,  be  useful  for  the  student  to  verify  that  principle 
when  convenient. 


FORMULAE  FOB  OBLIQUE  TRIANGLES.  59 

which  substituted  in  the  preceding  equation  gives 

a2  =  b2  +  <?  —  2  be  cos  A  (221) 

as  was  to  be  proved.     We  have  in  the  same  way 

62  =  a2  +  c2  —  2  ac  cos  E  (222) 

C2  =  a2  +  b2  -  2  ab  cos  G  (223) 

120.  The  same  result  is  obtained  from  the  following  equations  (which  are  evident 
from  Fig.  16,  where  c  =  A  P  -f-  P  B) 

a  =  6  cos  C  -\-  c  cos  B 


b  =  c  cos  A  +  a  cos  C 
c  =  a  cos  B  -{-  b  cos  -4 

From  the  first  of  these 

c  cos  B  =  a  —  6  cos  C 

whence  c1  cos*  B  =  a2  —  2  ab  cos  (7  +  6s  cos1  (7 

and  from  (218),         c»  sin2  JB  =  62  sin2  C 

the  sum  of  which  two  equations  is,  by  (13), 


(224) 


121.  From  (221)  we  find 

A2  _J_  ,2  _  n2 

™A=^rL  <225> 

by  which  an  angle  is  found  when  the  three  sides  are  given  ;  but  to 
adapt  it  for  convenient  computation  by  logarithms,  the  following 
transformations  are  necessary  : 

Subtract  both  members  from  unity;  then 

2  be  —  b2  -  <?  +  a2       a2  —  (6  —  c)2 
1  —  cos  ^4  =  -  —  =  - 

2  be  2  be 

But,  by  (139),  making  x  =  A,  we  have 
1  —  cos  A  =  2  sin2  1  A 

Also,  the  numerator  of  the  second  member  being  the  difference  of 
two  squares  may  be  resolved  into  two  factors,  viz.  :  the  sum  and  the 
difference  of  a  and  b  —  c  ;  thus, 


a2-(6  —  c)2  —  [a-(6-c)]X[a  +  (6-c)]  —  (a  —  6  +  c)(a+6  —  c) 
Substituting  these  values  in  the  above  equation  and  dividing  by  2 

sin2M  =  (a~6  +  C)iL(a  +  6~-^  (226) 

4  be 


60 


PLANE  TRIGONOMETRY. 


This  may  be  simplified  by  representing  the  half  sum  of  the  three 
sides  by  s,  or  by  putting 

a+  6  +  c  =  2s 
whence 

a  —  b+c=a+b+c—  2b  =  2s-2b  =  2(s~b) 
a-\-b  —  c  =  a  +  b-\-  c  —  2  c  =  2  s  —  2  c  =  2  (s  —  c) 

which  substituted  in  (226)  give 


•  2  i  (s  —  b)  (s  —  c) 

sm^  i  A  =  *• — 

be 

In  the  same  manner  we  should  find  from  (222)  and  (223) 


sin2  i  B  = 


ac  ab 

122.  Add  both  members  of  (225)  to  unity;  then 

1  +  cos  A  = 
But  by  (138) 


(227) 


(228) 


2  be 


2  be 


1  -f  cos  A  =  2  cos2  1^  A 

Also,         (6  +  c)2  —  a2  =  (b  -f  c  -f  a)  (6  +  c  —  a) 
therefore 


cos 
cos  • 


Substituting  s  in  the  numerator  as  in  the  preceding  article 


s  (s  —  a) 

cos2  i  A  =  — ^— « 

6c 


and  therefore  also 

cos2  ^  £  = 


oc 


2  i   /^        s  (t9  —  c) 

cos24  C—  -A—  z  —  * 


123.  Dividing  (227)  by  (229),  we  have,  by  (14) 


s  (s  —  a) 


(229) 
(230) 
(231 


and  in  the  same  manner 


tan2  1  B  = 


(s-b) 


tenUC=^  (23?) 

s  (s  —  c) 


FORMULA  FOR  OBLIQUE  TRIANGLES. 


61 


124.  The  preceding  formulae  are  sufficient  for  the  resolution  of  all  the  usual  cases 
of  plane  triangles;  but  there  are  others  that  are  occasionally  useful.  From  (218)  we 
find,  by  (105),  (106)  and  (135), 


sin  A  +  sin  B  _  sin  }  (A  +  B)  cos  %  (A  —  E) 
sin  C  sin  £  C  cos  %  C 


a  —  b sin  A  —  sin  B cos  ^  (A  -f-  B)  sin  j  (A  —  B) 

c  sin  C  sin  |  C  cos  J  G* 


But  since  A  +  B  +  O=  180°, 

A  +  B  =  180°  —  C, 
sin  $(A  +  B)=cos$  C, 
by  means  of  which  we  find 


I  (A  +  B)  =  90°  —  }  C 
cos  }  (.4  +  5)  =  sin  J  C 


a  -|-  &  _  cos  j^  (^4  —  .B)  _  cos  £  (  A  —  B)  (233^ 

c  sin  J  C       ~~  tos  }  (  J.  +  5) 

a  —  b  _  sin  ^  (,4  —  B)  _  sin  $  (A  —  B)  (234) 

c  cos  i  C       ~*in}(A  +  B) 

The  quotient  of  (233)  divided  by  (234)  gives  (220). 

125.     Adding  unity  to  both  members  of  (233),  or  subtracting  it,  we  have,  by  (103) 
and  (104) 


a  +  b  -f  e  _  cos  $  (A  +  B)  -f-  c°s  \( 
c  cos  %  (A-{-  B} 


B}  __  2  cos  %  A  cos  fr  B 
sin  J  C 


—  c  __  cos 


—  ff)  —  cos  ^  (A  +  B)  _  2  sin  %  A  sin  fr  ^ 
c  cos  i  (  ^1  +  B)  sin  £  (7 

Similarly  from  (234)  we  find 

c  -f  o  —  b  __  sin  \  (A  -f  /?)  -4-  sin  ^  (^l  —  ff)  _  2  sin  $  A  cos  %  B 
c  sin  £  (,4-f  />')  cos  $  C 


_  sin  \  (A  +  B)  —  sin  ^  (.4  —  jB)  _  2  cos  \A  sin  ^  .5 
sin  $(A  +  B)  cos  i  (7 


Substituting  s  =  }  (a  +  6  -f  c),  these  equations  become 


s_  _  cos  ^  A  cos 
c  sin  J  (7 

sin 


a  —  c  _  sn 

c  sin  £  (7 

»  —  6  _  sin  ^  A  cos  <fr  ^ 
c  cos  J  (7 

g  —  a  _  cos  $  A  sin  %  B 
c  cos  £  C 


(235) 

(236) 
(237) 

(238) 


62  PLANE  TKIGONOMETKY. 

From  these  equations  we  can  deduce  immediately  (227)  etc.  ;  for  example  exchang- 
ing c  for  6  in  (236),  we  have 

s  —  6  sin  ^  A  sin  ^  O 

6  sin  £  B 

which,  multiplied  by  (236)  gives  (227). 

126.  Four  times  the  product  of  (227)  and  (229)  is  by  (135) 

4 
sin*  A  =  — —  .  s  (s  —  a)  (s  —  b)  (s  —  c) 

whence 

be 
Exchanging  A  for  B  and  C  successively,  this  gives  also 

sin  B  =  —  i/  [s  (s  —  a)  (s  —  6)  (a  —  c)]  (240) 

ac 

sin  C=4l/ [«(•-«)  («-&)  («-«)]  (241) 

00 

In  these  equations  put  * 


then 


f>     Tr  O      "FT  O      TT 

2Ji  sin  5  =  ?^  sinC=?-f  (243) 

be  ac  ab 


The  quotient  of  the  first  of  these  divided  by  the  second  is 

sin  A ac a 

sin  B       be      b 

which  brings  us  back  to  the  theorem  of  Art.  117. 

127.  The  sum  of  A,  B  and  C  being  180°,  and  the  sum  of  $  A,  J  B  and  J  C  being 
90°,  we  have,  by  Arts.  85  and  86,  the  following  relations  among  the  angles  of  a  plane 
triangle. 

tan  A  +     tan  B  +     tan  0=  tan  A  tan  B  tan  C 
cot  i  A  +  cot  }  B  +  cot  \  C—  cot  J  ^4  cot  }  B  cot  }  C 
sin  J.  +     sin  1?  -|-     sin  C=  4  cos  J  .4  cos  i  5  cos  £  C 
sin  .4  +      sin  B  —      sin  0=4  sin  \  A  sin  j-  ./?  cos  J  C' 
cos  ^4  +  cos  B  -f-  cos  C  —  1  =  4  sin  J  A  sin  J  5  sin  J  (7 
cos  ^4  -f-  cos  B  —  cos  (7+  1  =  4  cos  \  A  cos  J  J?  sin  J  (7 

in  the  last  of  which  we  may  interchange  A,  B  and  C.     These  relations  may  be  substi- 
tuted in  the  equations  of  Art.  125. 

*  K  is  the  area  of  the  triangle.     See  Art.  148. 


FORMULJ2  FOR  OBLIQUE  TRIANGLES. 


63 


128.  The  following  equations  are  added  as  exercises. 
tanM   _s-b 


tan  J  B        a  —  a 

sin  (A  —  B}  _  (a  +  6)  (a  —  6) 
sin  (A  +  B)  c» 

tan  J  .4  tan  £  B  cot  £  (7=  ^          ^ 

JL 


oot  M  +  cot  }  J?  +  cot  }  C= 
sin  i  ^4  sin  J  B  sin  ^  C= 
cos  J  A  cos  J  5  cos  £  (7= 


—  —  • 
abcs 


tan 


ooc 

\B  tan  J  C= 


- 


—  (a2  —  61— 


-a4  —  6*  — 


CHAPTER    VIII. 


SOLUTION  OF  PLANE  OBLIQUE  TEIANGLES. 

FIG-18-  129.  CASE  I.     Given  two  angles  and  one 

side,  or  A,  B  and  a.     Fig.  18. 
To  find  the  third  angle.     We  have 

C=  180°  —  (A +  3) 
To  find  b  and  c.     We  apply  the  theorem  of  Art   117,  and  state 
the  proportions  thus  :  the  sine  of  the  angle  opposite  the  given  side  is 
to  the  sine  of  the  angle  opposite  the  required  side,  as  the  given  side 
is  to  the  required  side.     Thus  we  have 


whence 

and 

whence 


sin  A  :  sin  B  =  a  :  b 

,       a  sin  B  .     ~ 

6  =  — ; —    -  =  a  sin  B  cosec  A 
sin  A 

sin  A  :  sin  C=a  :  c 
a  sin  C 


c  = 


sin  A 


=  a  sin  C  cosec  A 


(244) 


(245) 


EXAMPLES. 

1.  Given  A  =  50°  38'  52",  5  =  60°  7'  25"  and  a  =  412-6708,  to 
find  C,  6  and  c. 

^  +  #=110°  46'  17" 
d=    69°  13' 43" 

By  (244).  By  (245). 

log  cosec  0-1 1 16730  log  cosec  O'l  1 16730 

log  sin  9-9380702 

log  sin  9-9708 129 
log  2-61 56037 


A  =  50°  38'  52" 
72  =  60°  7' 25" 
C  =  69°  13' 43" 
a  =  412-6708 


log  2-61 56037 

log  6  2-6653469 

6  =  462-7505 


log  c  2-6980896 
c=-498-9875 


fi4 


SOLUTION  OF  OBLIQUE  TRIANGLES.  65 

2.  Given  4  =  100°  16'  35",  £  =  25°   16'  13",  and  6  =  29-167, 
find  a  and  c. 

Am.  a  =  67-22857 
c=55-59178 

130.  CASE  I.  Given  A,  B  and  a.     Second  solution.    We  find  C=  180°  —  (^1  +  5)  ; 
then,  by  (233)  and  (234) 

(246) 


.. 
cos  J  (-B  +  C)  sin  }  A 

§-.-•.  !"Li-i*=£l=  a  .  Bi°t(J-0)  (247) 

sin  *  (-B  +  C)  cos  M 

wliich  give  the  sum  and  difference  of  the  required  sides;  adding  half  the  difference  to 
half  the  sum,  we  find  the  greater  side,  and  subtracting  half  the  difference  from  half  the 
sum,  we  find  the  less  side. 

131.  CASE  I.  Given  A,  B  and  a.  Third  Solution.  When  A  and  B  are  nearly  equal, 
and  great  accuracy  is  desired,  we  may  compute  the  difference  between  o  and  6;  for  we 
have,  from  (244), 

,  a  sin  B  sin  A  —  sin  B 

a  —  o  =  a  --      —  =  a  .  --  :  —  -  - 
sin  A  sin  A 

or 

a  —  b  =  2  a  Cos  KA  +  B}  sin  ^A  ~  B)  (248) 

sin  A 

EXAMPLE.  Given  A  =  35°  40'  12"-3,  B  =  35°  37'  48"-6,  and  a  =  26246-948. 

A  =  35°  40'  12"-3  log  cosec  0-2342442  0'23424 

B  =  35°  37'  48//-6  log  2    0'3010300  0'30103 

l(A+B)  =  35°  39'    0"-5  log  cos  9-9098720  9*90987 

I  (A  —  B)=   0°    V  ll"-9  log  sin  6-5423038  6.54230 

a  =  26246-948  log  4-4190788  4-41908 

a_  ft—       25-499  log  1-4065288  1-40652 
6  =  26221-449 

One  of  the  advantages  of  this  process  is,  that  a  —  b  may  be  found  with  sufficient 
accuracy  with  five-figure  tables,  as  in  the  second  column  of  logarithms  above.  If  a 
had  been  given  to  ten  figures  instead  of  eight,  we  should  still  have  been  able  with  the 
seven-figure  logs,  to  find  a  —  6  to  seven  figures,  and  therefore  b  to  ten  figures,  which 
could  not  be  done  by  the  ordinary  methods  without  ten-figure  tables. 

132.  CASE  II.  Given  two  sides  and  an  angle  opposite  one  of  them, 
or,  a,  6  and  A. 

To  find  B.  To  find  the  angle  opposite  the  other  given  side,  we 
apply  Art.  117,  and  state  the  proportion  thus:  the  side  opposite  the 
given  angle  is  to  the  side  opposite  the  required  angle  as  the  sine  of 
the  given  angle  is  to  the  sine  of  the  required  angle.  Thus,  with  the 
present  data,  we  have 

a  :  b  =  sin  A  :  sin  B  whence  sin  B  =  -  (249) 

a 

9  r  2 


66  PLANE  TRIGONOMETRY. 

To  find  C.  We  have  C=  180°  —  (A  +  B) 

To  find  c.  Having  found  C,  we  now  have  the  data  of  Case  I. ; 
therefore,  by  (245) 

c  =  a  sin  C  cosec  A  (250) 

133.  It  is  shown  in  geometry  that  when  two 
sides  and  an  angle  opposite  one  of  them  are 
given,  there  may  be  constructed  two  triangles, 
as  in  Fig.  19,  whenever   the   given  angle  is 
acute  and  the  given  side  opposite  to  it  is  les* 

than  the  other  given  side.  In  one  of  them,  the  required  angle  B  is 
acute,  and  in  the  other  it  is  obtuse,  and  the  two  values  are  supple- 
ments of  each  other ;  for 

B  =  B  B'  C=  180°  —  A  B'  C 

These  two  values  of  B  are  given  in  the  trigonometric  solution  by 
the  consideration  that  sin  B  found  by  (249)  is  at  once  the  sine  of  an 
acute  angle,  and  the  sine  of  its  supplement,  Art.  39. 

In  general,  when  an  angle  is  determined  only 
by  its  sine,  it  admits  of  two  values,  supplements 
of  each  other,  unless  the  conditions  of  the 
problem  are  such  as  to  exclude  one  of  these 
values.  In  the  present  case,  the  obtuse  value 
of  B  is  excluded  when  a  is  greater  than  6,  and 
there  is  but  one  triangle  whether  A  is  acute  or 
obtuse,  as  in  Fig.  20. 

134.  If  the  given  parts  were  such  that 

a  =  b  sin  A 

a  would  be  equal  to  the  perpendicular  from  C  upon  the  side  c,  and 
we  should  have  but  one  solution,  namely,  a  right  triangle,  B  and  its 
supplement  both  being  90°. 

135.  If  the  given  parts  were  such  that 

a  <C  b  sin  A 

a  would  be  less  than  the  perpendicular  from  C  and  the  problem 
would  be  impossible.  It  would  also  be  impossible  if  a  <  b  while 
A  >  90°. 

136.  When  there  are  two  solutions,  represent  the  two  values  of  B 
by  B'  and  B",  then  the  two  values  of  Cwill  be 

C'  =  180°  —  (A  +  B'}  ==  180°  —  B'  —  A  =  B"  —  A      (251) 
C"  =  180°  —  (A  +  B"}  =  180°  —  B"  -  A  =£'  —  A      (252) 


SOLUTION  OF  OBLIQUE  TRIANGLES.  67 

and  the  two  values  of  c  will  be 

c'  =  a  sin  C"  cosec  A          c"  =  a  sin  C"  cosec  A          (253) 

EXAMPLES. 

1.  Given  «=  31-23879,  6  =  49-001  17  and;!  =  32°  18';  find  5,  C 
and  c. 

«  =  31-23879  ar.  co.  log  8-5053058 

b  =  49-001  1  7  log  1-6902064 

A  =     32°  18'  log  sin  9-7278277 

B'  =     56°  56'  56"-3         log  sin  9-9233399 
.#"=123°    3'    3"-7 
C'  =     90°  45'    3"-7         log  sin  9-9999627 

C"  =    24°  38'  56"-3  log  sin  9-6201962 

log  cosec  A  0-2721723  0-2721723 

log  a  1-4946942  1-4946942 

log  c'  1-7668292  log  c"  1-3870627 

c'  =  58-45601  c"  =  24-38163 

Ana.  B  =  56°  56'  56"-3    *|         f    £  =  123°    3'    3"-7 
C=  90°  45'    3"-7     I  or  \     (7=     24°  38'  56"-3 

c=     24-38163 


2.  Given  a  =  -051234,  b  =  -042356,  A  =  55°  ;  find  B,  C  and  c. 

Ans.  B  =6=  42°  37'  32"-7 
C=  82°  22'  27"-3 
c=  .06199202 


(  B  =  97°  45'  24"-3 
C=27°  14'  35"-7 


3.  Given  a  =  -042356,  b  =  -051234,  yl  =  55°  ;  find  B,  Cand  c. 

4n*.  ^  =  82°  14'  35"-7      "| 

C=42°  45' 24"-3       lor 

c=     -03510331  (^    c  = -02366993 

4.  Given  a  =  40,  6  =  50,  A  =  53°  7'  48"-4 ;  find  A 

Ans.  jB  =  90°. 

5.  Given  a  =  40,  6  =  50,  A  =  60°  ;  solve  the  triangle. 

Ans.  Impossible. 

6.  Given  6  =  40,  c  =  50,  B  =  100°  ;  solve  the  triangle. 

Ans.  Impossible. 


68  PLANE  TRIGONOMETRY. 

137.  CASE  II.    Given  o,  6  and  A.    Second  Solution.  We 
may  solve,  separately,  the  two  right  triangles  AP  0,  B  PC, 
Fig.  21,  which  is  a  convenient  method  when  there  are 
two  solutions.     We  first  find  B  by  (249)  ;  then  we  have 
P  AP  =  bcosA,  £P=acoaB 

and  c  =  A  P  +  B  P 

The  cosine  of  the  obtuse  value  of  B  is  negative,  (Art.  39),  so  that  B  P  is  then  nega- 
tive, and  we  have  the  two  values  of  c  from  the  formula 

c=AP±BP 

There  will  be  but  one  solution,  if  B  P  >  A  P,  for  c  cannot  be  negative. 
138.  CASE  III.   Given  two  sides  and  the  included  angle,  or  a,  6  and  C. 
To  jind  A  and  B.     We  have  first 

A  +  B  ==  180°  —  C 
H^  +  -B)  =  90°-iC 

from  which  we  next  find  the  half  difference  of  A  and  B  by  the 
theorem  of  Art.  118,  which  gives 

a  +  b  :  a  —  b  =  tan  £  (A  +  B)  :  tan  £  (A  —  B) 


tan  4  (4  —  .B)  =        -    tan£(.A+  J3)  =    --    cot*  C       (254) 
a  -f-  o  a  -f-  6 

The  half  difference  added  to  the  half  sum  gives  the  greater  angle, 
(opposite  to  the  greater  given  side),  and  the  half  difference  subtracted 
from  the  half  sum  gives  the  less  angle. 

To  find  c.     We  have  the  data  of  Case  I.,  and  therefore 

c  =  a  sin  (7  cosec  A  =  b  sin  C  cosec  B  (255) 

EXAMPLES. 

1.  Given  a  =  '062387,  6  =  -023475,  and  C=  110°  32'  ;  find  A,  B 
and  c. 

A  +  B  =  180°  —  C=  69°  28' 
a+b  =    -085862         ar.  co.  log  1-0661990 
a  —  b=-.    -038912  log  8-5900836 

B)=--    34°  44'  log  tan  9*8409174 


£•  (A  —  B)  =     17°  26'  33"  log  tan  9-4972000 
A=    52°  10'  33" 

B  =     17°  17'  27"  log  cosec  0-5269189 

C=  110°  32'  log  sin  9-9714931 

b  =  -023475  log  8-3706056 

c  =  -0739635  log  8-8690176 

Ans.  A  =  52°  10'  33" 
5=17°  17'  27" 
c  =  -0739635 


SOLUTION  OF  OBLIQUE  TRIANGLES.  63 

2.  Given  a  =  31-0005,  6  =  15-1101,  C=  10°  15'  ;  find  A  and  B. 

Ans.  ^  =  160°  17'  13"-7 
B  =      9°  27'  46"-3 

3.  Given  a  ==  2-463878,  6  =  9-021875  and  C=  74°  9'  4"-2  ;  find 
A  and  B. 

Ans.  A  =  15°  50'  55"-8 

5  =  90°    0'    0" 

4.  Given  b  =  15-1101,  c  =  31-0005,  A  =  10°  15'  ;  find  B  and  C. 

^ns.  J5  =      9°  27'  46"-3 
C=  160°  17'  13"-7 

139.  Having  found  A  and  B  as  above,  the  most  convenient  mode  of  finding  c  is  by 
(233)  or  (234),  which  give 


(257> 


for  we  have,  from  the  process  of  finding  A  and  B,  the  log.  of  a  -f-  b,  or  of  a  —  b,  and 
the  values  of  £  (.4  +  -B)  an(i  J  (^  —  -^)»  so  tnat  we  have  only  two  new  logs,  to  find, 
which  are  taken  out  at  the  same  opening  of  the  tables  with  the  tangents  of  J  (A  -\-  B) 
and  J  (A  —  B). 

140.  CASE  III.  Given  a,  6  and  (7.  Second  Solution.  When  a 
and  6  are  given  by  their  logarithms,  which  occurs  when  they  are 
deduced  by  a  logarithmic  process  from  other  data  (as,  for  example,  in 
the  computation  of  a  series  of  triangles  in  a  survey),  we  proceed  as 
follows.  Let  x  be  an  auxiliary  angle,  such  that 

tan*  =  7  (258) 

6 

an  assumption  always  admissible,  since  a  tangent  may  have  any  value 
from  0  to  oo  . 
We  deduce 

tan  x  —  1  _  a  —  b 

tan  x  -f-  1       a  +  6 

or  by  (1  52)  tan  (a;  -  45°)  =  ^—  ^ 

a  -f-  6 

which  substituted  in  (254)  gives 

tan  %  (A  —  B)  =  tan  (x  -  45°)  tan  %  (A  +  JB)  (259) 


70  PLANE  TRIGONOMETRY. 

We  find  x  from  (258)  and  employ  its  value  in  (259).  As  this 
method  does  not  require  the  preparation  of  a  -f-  b  and  a  —  6,  it  is 
quite  as  short  in  practice  as  (254). 

EXAMPLE. 

Given  log  a  =  8-7950941,  log  b  =  8-3706056,  and  C=  110°  32'. 
(Same  as  Ex.  1.  Art.  138.) 

log  a  =  8-7950941 

log  b  =  8-3706056 

x  =  69°  22'  46"-8       log  tan  0-4244885 

x  —  45°  =  24°  22'  46"-8       log  tan  9-6562825 
$(A  +  £)  =  34°  44'  log  tan  9-8409174 

$(A  —  B)  =  17°  26'  32"-9       log  tan  9-4971999 

141.  CASE  III.  Given  a,  b  and  C.  Third  Solution.  To  express  A  or  B  directly  in 
terms  of  the  data,  we  have,  from  (218)  and  (224) 

c  sin  A  —  a  sin  C 
c  cos  A  =  b  —  a  cos  C 
the  quotient  of  which  is 


and  in  the  same  manner 


tan  .4=      "sn^  (260) 

6  —  a  cos  C 


ton  3  =  (261) 

a  —  b  cos  C 


142.  CASE  III.     Given  a,  b  and  C.     Fourth  Solution.     To  find  c  directly  from  the 
data,  we  have,  by  (223) 


&  =  a*  +  6*  —  2  ab  cos  C 

which,  however,  is  not  adapted  for  logarithmic  computation.     It  may  be  adapted  as 
follows.     Substitute  by  (139) 

cos  (7=  1  —  2  sin*}  C 

then  c*  =  a'-f&s  —  2a6  +  4a6sin2  J  (7 

=  (a  —  6)*  +  4  ab  sin1  J  C 


(262) 

Let  x  be  an  auxiliary  angle,  such  that 

tan,          4o6Bin'}(7 
(a  -by 

siniC 
a  —  b 


SOLUTION  OF  OBLIQUE  TRIANGLES.  71 

then  the  radical    in  the   above  equation  becomes  \/  (1  +  tan*  x)  =  sec  x  ;  therefore, 

c=(a.  —  b)secx  (264) 

143.  We   may  also  adapt  (223)  for   logarithmic  computation  by  means  of  (138), 
which  gives 

cosC  =  —1  -f  2  cos2  J  C 
whence 

c»  =  o»  +  62  +  2  ab  —  4  aft  cos1  $  C 


(265) 

Let 

(266) 


then  the  radical  becomes  y  (1  —  sin1  x)  =  cos  z;  therefore, 

c  =  (a  +  6)  cos  z  (267) 

144.  It  is  to  be  observed,  that  the  supposition  (263)  is  always  possible,  jjnce  a  tan- 
gent may  have  any  value  between  0  and  oo  ,  and  therefore  an  angle  x  may  always 
be  found  having  any  given  number  as  its  tangent.  As  the  greatest  value  of  a  sine  is 
unity,  it  is  not  so  obvious  that  the  supposition  (266)  is  always  possible;  but  whatever 
the  values  of  a  and  6 

(a  —  6)1  ^  0 
therefore  (a  +  ft)1  >_  4  ab 

2l/~a5 
whence 

a  -f-  o    = 

therefore  the  second  member  of  (266)  is  never  greater  than  unity. 

EXAMPLE. 
Given  a  =  "062387,  6  =  -023475,  C=  110°  32';  (same  as  Ex.  1,  Art.  138) 

By  (266)  and  (267). 
a  =  -062387  log  8-7950941 

6  =  -023475  log  8-3706056 

2)7-1656997 


=  8-5828499 

a  4-  b  =  -085862  ar.  co.  log  1-0661990  log  8'9338010 

£  (7=  55°  16'  1.  cos  9-7556902 

log  2  0-3010300 

1.  sin  x  9-7057691  1.  cos  x  9'9352161 

c  =  -07396344  log  S'8690171 


72 


PLANE  TRIGONOMETRY. 


145.  CASE  IV.      Given  the  three  sides,  or  a,  b  and  c. 
To  find  A,  B  and  C.     We  have  from  (227)  and  (228X 


sin  i  A  =  ~  '  '  * "- ' 


.     ,   „          //(«-a)(«  — 
sin  |  .S  =  A/  I  ^ — 

»    \  CLG 

,mlc=  l/(s~a^s- 

or  by  (229)  and  (230) 


(268) 


cos  i  C'  = 


(269) 


or  by  (231)  and  (232) 


(270) 


In  these  formula  s  =  ^  (a  +  b  -\-  c).  Either  of  these  three  meth- 
ods may,  in  general,  be  employed,  but  (268)  is  to  be  preferred  when 
the  half  angle  is  less  than  45°,  and  (269)  when  the  half  angle  is 
more  than  45°.*  When  all  the  angles  are  required,  (270)  will  be 
the  simplest,  as  it  requires  but  four  different  logs,  to  be  taken  from 
the  tables.  It  is  accurate  for  all  values  of  the  angle. 


*  See  Art.  112. 


SOLUTION  OF  OBLIQUE  TRIANGLES.  73 

EXAMPLES. 

1.  Given  a  =  10,  6  =  12,  c  =  14  ;  find  the  angles. 

By  (268). 

a  =  10  ar  co  1  9-0000000    ar  co  1  9-0000000 

6=12  ar  co  1  8-9208188  ar  co  1  8-9208188 

c  =  14  ar  co  1  8*8538720    ar  co  1  8-8538720 
2  «  =  36 

8  =  18 

a  —  a=    8  log  0-9030900          log  0-9030900 

s  —  b=    6        log  0-7781513  log  0-7781513 

8  —  c=    4       log  0-6020600         log  Q-6020600  _ 

2)9-1549021  2)9-3590220  2)9-6020601 

log  sines          £-49-5774510        £  59-6795110         £(79-8010300 
|^t  =  22°12'27"-6    £  B  =  28°  33'  39"-0    \  (7=  39°  13'  53"-5 
A  =  44°  24'  55"-2       5  =  57°    7'  18"-0       0=  78°  27'  47"-0 

Verification.  A  +  5  +  (7=  180° 

2.  Given  a  =  -8706,  6  =  -0916,  c  =  -7902  ;  find  the  angles. 

Am.  A  =  149°  49'  0"-4 
B=  3°  1'  56"-2 
C=  27°  9'  3"-4 

3.  Given  a  =  -5123864,  6  =  -3538971,  c  =  -3090507  ;  find  C. 

Ans.  C=36°  18'  10"-2 

146.     The  computation  by  (270),  when  all  the  angles  are  required,  will  be  much 
facilitated  by  the  introduction  of  an  auxiliary  quantity* 


(271) 
from  which  we  find  by  (270) 

tan  J  A  =  -^—,  tan  J  B  =  —?—,  tan  }  0=      r  (272) 

s  —  a  s  —  6  8  —  c 


*This  quantity  r  is  the  radius  of  the  inscribed  circle.     See  (289). 
10  Q 


74  PLANE  TRIGONOMETRY. 

EXAMPLE.     Given  o  =  6053,  b  =  4082,  c  =  7068.     We  find 

s  =  8601-5  ar.  co.  log  6  0654258 

s  —  a  =  2548-5  log  3'4062846     , 

a  —  b  =  4519-5  log  3  6550904 

<  —  e  —  1 533-5  log  3-1856838 

2)6-3124846 
logr          =    3-1562423 

J  A  =  29°  20"  54"'47    •  log  tan  *  A  =  log  — T—     =     9'7499577 

s —  a 

J  J5  =  17°  35'  31"-70  log  tan  }  B  =  log  -£-     =    9'5011519 

s  —  b 

J  C=  43°  3'  33"'83  log  tan  J  C  =  log  — —     =    9'9705585 

Verification.    90°  0'    0"'00 

147.  The  case  where  the  three  sides  are  given  is  some- 
times solved  as  follows.  From  C,  Fig.  22,  draw  CP 
perp.  to  c.  Then 


£-  -*« 

the  difference  of  which  is 

A  C*  —  B  C"  =  A  P*  —  B  P1 

or  (4C+.BC)  (40— £C)  =  MP+£P)  (4P- 

and  if  .4  P  —  B  P=  d,  this  equation  gives 

d(o  -f-  a)  (o  —  a)  /«>7M\ 

=  * — • — •-* '  (***) 

c 

Then,  since  A  P-f  BP  =  c,  and  ^4  P—BP~d,  we  have 

^P=J(c  +  d),  ^P=J(c  — d)  (274) 

and  in  the  right  triangles  A  CP,  B  CP 

cosA  =  ^f,  c*xB=S-p-  (275) 

o  a 

30  that  (273),  (274)  and  (275)  solve  the  problem.     When  d  >  c,  B  P  is  negative,  cos  B 
is  negative,  and  B  is  an  obtuse  angle,  (Art.  39). 

AREA  OF  A  PLANE  TRIANGLE. 

148.  Representing  the  area  by  K,  and  the  perpendicular  CP,  Fig.  22,  by  p,  we  have, 
by  geometry, 

JT=Jcp  (276) 

In  the  triangle  A  C  P,  we  have  p  =  b  sin  A,  whence 

K—  J  be  sin  A  =  be  sin  J  A  cos  }  A  (277) 

by  which  the  area  is  compute*!  from  two  sides  and  the  included  angle. 

Substituting  in  (277)  the  values  of  sin  J  A  and  COB  }  A  by  (268)  and  (269), 

K^^/[s(s  —  a)  (s  —  b)  («-c)]  (278) 

by  which  the  area  is  computed  from  the  three  sides. 


CHAPTER    IX. 

MISCELLANEOUS  PROBLEMS  RELATING  TO  PLANE  TRIANGLES. 

149.     In  a  given  plane  triangle,  to  find  the  perpendicular  from  one  of  the  angles  upon  the 
opposite  side. 
Let  p  be  the  perpendicular  from  C  upon  c.    We  have 

p  =  b  sin  A  (279) 

or  by  (239)  and  (278), 


the  expression  for  p  in  terms  of  the  three  sides,  where  s  =  J  (a  -f-  6  -f-  c)  and  JT  is  the 
area  of  the  triangle. 

If  we  substitute  in  (279)  sin  A  =  -  sin  C,  it  becomes 


c 


=  -  sin  C  (281) 


or,  if  we  substitute  the  value  of  b  =  c  — — 

sin  C 


sin  A  sin  J? sin  A  sin  J? 

sin  C  s 


When  the  triangle  is  right-angled  at  C,  (282)  becomes 

p  =  c  sin  A  cos  A  —  -  sin  2  A 
2 

the  expression  for  the  perpendicular  upon  the  hypotenuse. 

150.  If  p't  p//,  p///,  denote  the  perpendiculars  upon  the  sides  a,  b,  e  respectively, 
we  have  from  (280) 


_J__L     1      i      1     _«  +  M-«_   » 
h     "  "~  '"    ~ 


2  A' 

76 


76  PLANE  TRIGONOMETRY. 

151.  To  find  the  radius  of  the  circle  circumscribed  about  a  plane  triangle. 

F10-  23-  The  center  0  of  the  circle,  Fig.  23,  lies  in  the  perpen- 

dicular erected  from  the  middle  point  of  one  of  the  sides,  as 
A  B.     Let  the  radius  =  R.     We  have,  by  geometry, 


and  in  the  triangle  A  0  D, 
'B 

•_  n       AD 


AO      2R 

whence  R  =  — 1—  (284) 

2  sin  O  V       ' 


Substituting  the  value  of  sin  C  from  (241), 

r>  abc  abc 


K  (285) 


From  (229)  and  (230),  we  easily  find 

cos  J4  MS  1  .B  cos  J  0= 


abc 
which  combined  with  (285)  gives 

jj  — &  (ftofi} 

4  cos  £  A  cos  J  B  cos  $  C 

152.  To  find  the  radius  of  the  circle  inscribed  in  a  plane  triangle. 

FlG- 24-  The  required  center  0,  Fig.  24,  is  in  the  inter- 

section of  the  three  lines  bisecting  the  angles,  and 
each  of  the  perpendiculars  0  D,  0  E,  OF,  is 
equal  to  the  required  radius  =  r.  The  value  of 
O  F  in  terms  of  A  B  =  c,  0  A  B  —  J  A,  and 

sin  \  A  sin  ^  B sin  fr  A  sin  |  B  (9K7\ 

an}(A+B)  ~  cos  }  G 

This  is  reduced  by  means  of  (236)  to 

r  =  (s  —  eJtanJC'  (288) 

Substituting  the  value  of  tan  J  C, 

r=J((s  —  a)  (*  —  b">  (*—  &\=K  (289) 

This  is  reduced  by  means  of  (231)  and  (232)  to 

r  =  s  tan  J  A  tan  J  J?  tan  }  C  (290) 


PROBLEMS  RELATING  TO  PLANE  TRIANGLES. 


77 


153.  Besides  the  inscribed  circle,  strictly  so  called,  there  are  three  other  circles 
that  touch  the  three  sides,  (or  sides  produced),  and  are  exterior  to  the  triangle,  as 
in  Fig.  25.  These  have  been  named  escribed  circles.  Their  centers  are  found  geomet- 


rically, by  bisecting  the  exterior  angles  £CC/,  CBB/,  etc.  Designate  the  centers 
of  the  circles  lying  within  the  angles  A,  B,  and  C  respectively,  by  0/,  O",  and  0"', 
and  their  radii  by  r',  r//,  r///.  We  find  the  perpendiculars  from  Of}  etc.,  upon  B  (7, 
etc.  by  (282)  to  be 


/  _       cos  \  B  cos  ^  O cos  \  B  cos  \  C 

'  sin  %(B-\-C)  cos  $  A 


}•       (291) 


tff cos  \  A  cos  \  B cos 

sin  ^  (.4  -f-  B) 

By  means  of  (235)  we  reduce  those  values  to 


r'  —  s  tan  \  A, 


-=  s  tan  j  B, 


cos  \  O 
r"f  =  s  tan  }  C 


Substituting  the  values  of  tan  £  A,  etc. 

r/  =    /  («(«-6) 
\\          «  — 


c)\  _. 


r'"  = 


(s  —  o)  (s  —  6)^  _ 

8  —  C 


(292) 


(293) 


78 


PLANE  TRIGONOMETRY. 


Also,  by  means  of  (236)  applied  successively  to  a,  b  and  e,  we  may  reduce  (291)  to 
the  following  : 

r'  —  («  —  o)  tan  J  A  cot  $  B  cot  J  C 
r"  =  (s  —  b)  cot  i  ^4  tan  $  J?  cot  $  C 


=  (s  —  c)  cot  J  /I  cot  $  5  tan  $  C 


(294) 


154.  Relations  between  the  radii  of  the  circumscribed,  inscribed,  and  thret  escribed  circles 
of  the  preceding  article,  and  the  three  perpendiculars  from  the  angles  upon  the  opposite 
sides. 

The  four  equations  of  (289)  and  (293)  give 


s  (•  —  a)  (s  -  b)  (s  -  c)        K* 
Dividing  this  successively  by  r1,  r'1,  etc. 

^~^-    =*'  ^9^=(8-a)' 

r  r'  r"'    _  r  r'  r" 


(295) 


(296) 


Again,  we  have,  (Art.  127), 

tan  }A  tan  £  B  +  tan  $  A  tan  \  C  +  tan  £  5  tan  \  C  —  1 
and  substituting  in  this  the  value  of  the  tangents  from  (292) 

r/  r//  _j_  r/  r/ 


1,1,1  1 
— T-H — 77-  +  -777  =  - 
r  r/x  rw  r 


(297) 


From  (292)  we  find 


tan  \A  _    T 
tan  \  B  ~~  r 


r"  tan  }  A  —  r'  tan  }  B 


from  which  it  follows  that  in  Fig.  25,  the  distances  A  D  and  B  D',  are  equal  (D,  D' 
being  the  points  of  contact  of  the  circles  0',  O'f  with  A  B  produced),  and  therefore 
B  D  =  A  D'.     Other  curious  geometrical  properties  may  be  traced  with  the  aid  of 
our  equations. 
From  (284), 

n  _  __  a  _  _  b  __  _  _  c  _ 

4  sin  J  A  cos  J  A        4  sin  J  B  cos  £  B        4  sin  J  C  cos  J  (7 

which  combined  with  (287)  and  (291)  give,  by  Art.  127, 

=  4  sin  J  >4  sin  j  B  sin  j  (7  =       cos  -4  -f  cos  -B  -|-  cos  C  —  1 


R 


-. 
R 

r" 
-—-  =  4  cos  }  A  sin  J  5  cos  f 

A 


-~ 

JB 


cos  ^4  —  cos  J5  -f-  cos  (7  -}-  1 
cos^l  +  cos-B  —  cos(7+l 


(298) 


PROBLEMS  RELATING  TO  PLANE  TRIANGLES.  79 

Changing  the  signs  of  the  first  of  these  equations,  the  sum  of  the  four  is 

r/  +  r//  +  r///"^-r   -  4,  B  =  t(r'  +  r"  +  »"'  -  r)          (299) 

Finally,  if  p',  p",  px//  denote  the  three  perpendiculars  from  the  angles  upon  the 
sides  a,  6,  c  respectively,  we  have  by  (283),  (289)  and  (297)  the  following  relation: 


JL  -4-  J_  4-    1     —  -L  4.  JL  4.  _A_  =  L 
•gf  *  p"  ^  $'"       *•*'-    r"'        r 


(299*} 


155.  To  find  the  distance  between  the  centers  of  tlie  circumscribed  and  inscribed  circles.* 


Let  P,  Fig.  26,  be  the  center  of  the  circumscribed, 
and  O,  that  of  the  inscribed  circle.  Put  P  O  =  D. 
By  Arts.  151  and  152, 


whence 


FIG.  26. 


and  by  (221) 


r>_ 

r> 


therefore 


A1-  —g 

-2PA.  OAcosPAO 


_»f  __  2-RrcoaH-B-C') 
sin2  \  A  sin  \  A 

_  4  Rr  sin  £  J?  sin  J  (7 


.        — 
sin4  5  vl 


(J  + 


(300) 


156.  Let  PO/  =  D/,  Fig.  26,  O7  being  the  center  of  the  escribed  circle  lying 
within  the  angle  A.  If  r'  =  radius  of  this  circle,  we  have,  as  in  the  preceding 
article 


n/z  —  PS  _u       r/t  2  Rr'  cos  \  (B—G 

-  ~T-'J1J  l      j 

j 


sin 

4  Rr'  cos  £  JS  cos  j  (7 
sin     J 


=  J?  4-  2  Kir 


(301) 


*  Hymers'  Trig.  Appendix,  Art.  58. 


80  PLANE  TRIGONOMETRY. 

the  expressions  for  the  distances  of  the  centers  of  the  three  escribed  circles  from  that 
of  the  circumscribed  circle. 
The  sum  of  (300)  and  (301)  gives  by  (299) 

Z>2  +  D'J  +  iyn  +  D"n  =  12  IP  (302) 

157.  Given  two  sides  of  a  plane  triangle  and  the  difference  of  their  opposite  angles,  (or 
a,  6,  and  A  —  £).  to  solve  the  triangle. 

We  have  \  (A  +  B)  directly  from  (220),  which  also  solves  the  case  where  two 
angles  and  the  sum  or  difference  of  two  sides  are  given. 

158.  Given  the  angles  and  the  sum  of  the  sides,  (or  A,  B,   C,  and  a4-6-f-c  =  2s). 
By  (235) 


cos  £  A  cos  J  B 

and  a  and  b  are  found  by  similar  formulae. 

159.  Given  one  angle,  the  opposite  side,  and  the  sum  of  the  squares  of  the  other  two  sidei, 
(or  C,  c,  and  a'  +  b3  =  e2). 

In  the  identical  equations 

(a  +  6)*  =  e2  -f  2  oi,  (a  —  b)t  =  es  —  2ab 

substitute  the  value  of  2  ab  given  by  (223),  namely, 

e1  —  c1 


2ab  = 


cos  C 


we  find  (a  +  6)»  =  e1  +  —-,     (a  -  6)»  =  *  -      = 

cos  (7  cos  C 

which  determine  a  -f-  b,  and  a  —  b,  and  therefore  a  and  b. 
To  compute  these  equations  by  logarithms,  let 


then  (a  +  6)*  =  e2  +  f,  (a  —  b)t 


(303) 


that  is  a  +  b  is  the  hypotenuse  of  a  right  triangle  whose  sides  are  e  and  g  ;  and  a  —  b 
is  one  side  of  a  right  triangle  whose  other  side  is  g,  and  whose  hypotenuse  is  e.  Let 
the  angle  opposite  g  be  denoted  by  x  in  the  first  triangle  and  by  x'  in  the  second,  then 
by  the  formulae  of  right  triangles 


tan  x  =  -* 
e 


sm  z'  =  •*  o  —  o  =  e  cos  z' 

e 


(304) 


so  that  the  problem  is  solved  by  logarithms  by  finding  log  g  from  (303)  and  employing 
its  value  in  (304). 

The  above  may  serve  as  an  example  of  a  geometrical  method  of  introducing  the 
auxiliary  quantities,  which  is  occasionally  useful.  The  analytical  process  in  the 
present  instance  is  similar  to  that  of  Art.  143;  thus 


PROBLEMS  RELATING  TO  PLANE  TRIANGLES.       81 

therefore  if  tan  x  =  -2  we  have  -»/  (  1  -f-      i  =  sec  z 

e  \  \         e1/ 

and  if  sin  xr  =  -^  we  have  -*  I  1  1  —   «  I  =  cos  a/ 

e  \  \         e1) 

whence  the  same  formulae  as  before. 

1  60.    Given  an  angle,  its  opposite  side,  and  the  difference  of  the  squares  of  the  other  two 
sides,  (or  C,  c,  and  aa  —  62  =/2). 

We  have  by  multiplying  (233)  by  (234) 

ain  (A  —  B)  _  olmi2  _  /* 
sin  0  #  # 

sin  (A  —  B)  —  £-  sin  C 
c1 

whence  A  —  B,  and  since  A  +  B  —  180°  —  C,  the  angles  are  determined.  There 
will  be  two  solutions  given  by  sin  (  A  —  B)  except  where  the  obtuse  value  of  A  —  Ji 
is  greater  than  A  -\-  B. 

161.  Given  the  three  perpendiculars  from  the  three  angles  upon  the  opposite  sides. 

Denote  the  perps.  upon  a,  b  and  c  respectively  by  a/,  b'  and  cx,  and  let 

" 


If  k  =  2  area  of  the  triangle 

aa'  =  bbf  =  cc'  =  k 
and  therefore 

a  =  a"  k,        b  =  b"  k,        c 

Substituting  these  values  of  a,  6  and  c  in  (225),  (227),  etc. 


, 


6"  e" 


in  which  2  s"  =  a"  +  6r/  +  c/r. 

162.  G-iwcw  <Ae  radii  of  the  circumscribed  and  inscribed  circles,  and  the  perpendicular 
from  one  of  the  angles  upon  the  opposite  side,  to  solve  the  triangle. 

Let  c  be  the  side  to  which  the  perpendicular  (p)  is  drawn.  We  have  found  for  R,  r 
and  p  the  expressions 


2  sin  C       2  sin  (A  +  B)      4  sin  }  (A  +  B)  cos  %(A-\-B) 
.   sin  £  -4  sin  i  B 


sin  A  sin^B 

'  sin  (4  +  B) 


11 


82  PLANE  TRIGONOMETRY. 

Eliminating  c  we  have 

^  =  sin  A  sin  B  (m) 

-^  =  4  cos  J  (4  -f  B)  sin  J  4  sin  }  B  (n) 

from  which  two  equations  .4  and  B  are  to  be  found.     Developing  cos  J  (.4  -f-  B), 
(n)  becomes 

^  -=  4  sin  J  .4  cos  J  ^4  sin  $  .B  cos  J  .B  —  4  sin1  J  ^4  sin*  £  B 
R 

=  sin  ^4  sin  B  —  4  sin*  £  .4  sin1  J  £ 
which  subtracted  from  (m)  gives 

£==£?  =  4  sin1  M  sin1  }  5  (0) 

Dividing  the  square  of  (m)  by  (o),  we  find 


whence 

sin  I  A  sin  }  J5  =  J  f  £=2r\  = 
\\SEJ 


cos        cos  J  5  = 


2l/[2J?(p-2r)] 
The  difference  and  sum  of  these  two  equations  give 

cos  J  (A  +  B)  =  -  r-  -  . 
VP*(j>-2r)] 

[     (305) 

cos  }  (A  —  B)  =  -  2^zl  - 
1/[2JR(p-2r)] 

which  determine  |  (A  -f  JS)  and  J  (.4  —  5)  and  therefore  A  and  .B.    The  sides  are 
then  found  by  the  formula 

c  =  2  R  sin  C 

Fl0-27-  163.  7ft  a  given  plane  triangle  ABC,  Fig.  27,  to  ^nd  a 

point  P  suth  that  the  three  lints  drawn  from  this  point  to  the 
angles  A,  B  and  C  shall  make,  given  angles  with  each  other. 

Let  the   given   angh-s   be  BPC=a        a.r\(\APC=(i 
and  the  required  angles  P  A  (7=  x  P  B  C  =  y 

_          The  sum  of  the  angles  of  the  quadrilateral  A  CB  P  is 


C=  360° 
whence  }  (x  +  y)  =  180°  —  J  (a  +  ft  -j-  C)  (306) 


PROBLEMS  RELATING  TO  PLANE  TRIANGLES. 
In  the  triangles  A  PC,  B  PC,  we  have 


83 


PC- 


b  sin .  x a  sin  y 

sin  /?          sin  a 


sin  x      a     sin  9 

—  —  =  -  .  - — «=  =m 

sin  y      o     sin  a 


from  which 


sin  x  -f-  sin  y  _  tan  |  (x  -f-  y)  _  OT  -p-  1 
sin  a;  —  sin  y      tan  J  (x  —  y)       TO  —  1 


tan  H*  -y)  = 


tan  *  (x  +  y) 


m  + 
To  compute  this  equation  by  logarithms,  let 

a  sin  8 
tan  7  =  m  =  — 

b  sin  a 

then  by  (152),         tan  J  (x  —  y)  =  tan  (7  —  45°)  tan  %  (x  +  y) 
so  that  the  angles  x  and  y  are  found  by  (306)  and  (307). 

164.  The  following  problems  are  proposed  as  exercises. 

In  a  plane  triangle  ABC  — 

1.  Given  c,  the  perp.  upon  c  =  p  and  a  -f-  b  =  m. 


(m  +  c)  (m  —  c) 


a  —  6  =  c  cos  x 
2  pc 


(m  +  c)  (TO  —  c) 
2.  Given  c,  the  perp.  upon  c  =  p,  and  a  —  b  =  n. 


tan7  x 


(c  -f  n)  (c  —  n) 


a  _[.  ft  —  c  gee  a: 


2  pc 


3.  Given  C,  c,  and  a&  =  <?. 


tan  x  =  — "  cos  J  C 

c 


a  -\-  b  =  c  sec  x 

sin  xf  =  — $  sin  %  C  a  —  6  =  c  cos  xf 

c 


4.  Given  (?,  the  perp.  from  C  =  p,  and  a  -}-  6  =  m. 

c  =  m  tan 

5.  Given  (7,  the  perp.  from  C  =  p,  and  a  —  6  =  n. 


tan  x  =  —  tan  J  C 
P 


tan  x  =•  —  cot 
P 

6,  Given  c,  C,  and  a  -f-  6  =  TO. 


cos  }  (4  —  B)  =  -  sin  }  (7 

C 


c  =  n  cot  J  x 


8111   #   (A  — 


(307) 


84  PLANE  TRIGONOMETRY. 

7.  Given  c,  A,  and  a  -f  6  =  m. 

tan  i  B  =  !lzr_c  cot  i  A 
m  -f-  c 

8.  Given  a  +  6  =  m,  the  perp.   upon  c  =  p,  and  the  difference  of  the  segments 
of  c  =  d. 


or  with  an  auxiliary  angle 

+  d)(m  —  d) 


•  *                   -4  »J 
sin1  x  = • — — f —  c  —  m  cos  x 


tan  i  (4  -f-  B)  =  ™L  sin  x  tan  x  tan  }  (A  —  B)  =  A  sin7  x 

2p  2p 

9.  Given  the  perimeter  =  2  s,  C,  and  the  perp.  from  C  =  p. 
tan*  z  =  -^  cot  J  (7 


10.  Given  c,  a  -{-  ft  =  ">  and  the  radius  of  the  inscribed  circle  =  r. 


m  —  e 
11.  Given  c,  a  —  b  =  n,  and  the  radius  of  the  inscribed  circle  =  r. 


4r" 
12.  Given  the  radii  >•',  r",  rfff,  of  the  three  escribed  circles.     (Arts.  153,  154.) 

r/2 

tan*  ft  A  =  _  - 
r/  rff  +  r/  r/// 


165.  Given  the  sides  of  a  quadrilateral  inscribed  in  a  circle,  to  find  its  angles  and  area. 
FIG. 28.  in   Fig.   28,  let   AB  =  a,  BC=b,  CD  =  c,    DA  —  d. 

-^  Let  2s  —  a  +  6-fc  +  d  and   ^T=area  of  A  BCD;    then 

from  the  triangles  ABC,  ADC,  observing  that  B  =  180°  —  D 
we  find 


06  +  cd  a6  -f  cei 


. 

—  c)  (s  —  d) 

=  v/  [(«  -  a)  (s  —  b)  (B  -c)(s-  d)]  (308) 


CHAPTER    X. 

SOLUTION  OF  CERTAIN  TRIGONOMETRIC  EQUATIONS  AND  OF  NU- 
MERICAL EQUATIONS  OF  THE  SECOND  AND  THIRD  DEGREES. 

166.  THE  solution  of  a  problem  in  which  the  unknown  quantity 
is  an  angle,  often  depends  upon  that  of  one  or  more  equations,  in- 
volving different  functions  of  the  angle,  which  cannot  be  reduced  by 
merely  algebraic  transformations.     We  shall  select  a  few  simple  ex- 
amples of  such  equations    from  among  those  that  most  frequently 
occur  in  astronomy. 

167.  To  find  z  from  the  equation 

sin  («  -f  z}  =  m  sin  z  (309) 

in  which  a  and  m  are  given.     We  have,  by  (119), 

sin  (a  -\-  z)  =  sin  a  sin  z  (cot  z  +  cot  a) 
which  becomes  identical  with  (309)  by  taking 

sin  a  (cot  z  -f  cot  a)  =  m 
whence  the  required  solution 

7?i 

cot2  =  -         -cot  a  (310) 

sin  a 

If  the  proposed  equation  were 

sin  (a  —  z)  — msinz  (31 1) 


we  should  find 


r*n 

cot2  =  -      -  +  cota  (312) 

sin  a 


Unless  2  is  limited  by  the  nature  of  the  problem  in  which  these 
equations  are  employed,  there  will  be  an  indefinite  number  of  solu- 
tions;  for  all  the  angles  z,  z  +  180°,  z -f-  360°,  z  +  540°,  etc.,  in 
general  all  the  angles  z  -\-  n  TT  have  the  same  cotangent.  [See  (68), 
(79).]  In  most  cases,  however,  we  consider  only  the  first  two  of 
these  solutions,  taking  the  values  of  z  always  less  than  360°. 

H  85 


86  PLANE  TRIGONOMETRY. 

Similar  remarks  apply  in  all  cases  where  an  angle  is  determined 
by  a  single  trigonometric  function  ;  but  if  the  problem  is  such  as  to 
give  the  values  of  two  functions  of  the  required  angle,  as  the  sine  and 
cosine,  the  solution  is  entirely  determinate  under  360°,  since  there 
cannot  be  two  different  angles  less  than  360°  that  have  the  same 
sine  and  cosine. 

168.  The  solution  of  the  preceding  article  requires  the  use  of  a 
table  of  natural  cotangents  ;  to  obtain  a  formula  adapted  for  logar- 
ithmic computation  entirely,  we  deduce  from  (309)  the  following 

sin  (a  -f-  2)  -|-  sin  z  _  m  -f  1 
sin  (a  +  z)  —  sin  2      m  —  1 

But  by  (109),  if  x  =  a  -f  z,  y  =  z,  we  have 

sin  (a  -f-  z)  -f-  sin  z  _  tan  (z  -|-  ^  a) 
sin  (a  -f  z)  —  sin  z  tan  |  a 

which  substituted  above,  gives 

t     ,    i    ,       m  -f  1 
tan  (z  -f-  IT  «)  =  -       -  tan  i  a 
m—1 

which  determines  z  -j-  ^  a,  whence  z  is  found  by  deducting  ^  a. 

The  computation  of  this  equation  is  facilitated  in  most  cases  by 
introducing  an  auxiliary  angle,  such  that 

tan  <p  =  m 

an  assumption  always  admissible,  since  while  the  angle  varies  from 
0  to  90°  the  tangent  varies  from  0  to  GO  ,  so  that  an  angle  (f  may 
always  be  found  having  any  given  number  as  its  tangent. 
We  have  then  by  (152), 


m—1       tan  <p  —  1 
and  the  preceding  solution  becomes 
tan^  =  wi,         tan  (z  +  £  «)  =  cot  (<p  —  45°)  tan  1  a  (313) 

The  logarithmic  solution  of  (311)  is  found  in  the  same  manner 
to  be 

tan  <p  =  m,  tan  (z  —  \  «)  =  cot  (y>  +  45°)  tan  £  a  (31  4) 


SOLUTION  OF  TRIGONOMETRIC  EQUATIONS.  87 

169.  To  find  zfrom  the  equation 

tan  (a  -f-  2)  =  m  tan  2  (315) 

We  deduce 

tan  (a  -f-  g)  H~  tan  2 m  +  1 

tan  (a  -f-  2)  —  tan  2       m  —  1 

so  that  by  (126)  and  (152)  the  solution  is 

tan  <p  =  m,  sin  (a  -f-  2  z)  =  cot  (^  —  45°)  sin  a        (316) 

170.  To  find  zfrom  the  equation 

tan  (a  -f  2)  tan  2  =  m,  (31 7) 

We  deduce 

1  -f  tan  («  -j~  z)  tan  2  _  1  -f  m 
1  —  tan  (a  -f  2)  tan  2       1  —  m 

so  that  by  (127)  and  (151)  the  solution  is 

tan^  =  m,  cos  (a  -f  2  2)  =  tan  (45°  —  ^>)  cos  a       (318) 

171.  To  find  z  from  the  equation 

sin  (a  ±  2)  sin  z  =  m  (319) 

By  (108)  we  find 

cos  a.  —  cos  (a  ±  2  z)  =  ±  2  sin  (a  ±  z)  sin  z  =  ±  2  m 

whence  cos  (a  rfc  2  z)  =  cos  a  =F  2  wi  (319  *) 

which  determines  a  =b  2  z,  and  hence  2  z. 

From  (319*)  we  have  four  values  of  a  ±  2z  between  0°  and  720°  ;  therefore,  four 
values  of  2  z  between  the  same  limits,  and  four  values  of  z  between  0°  and  360°. 

In  general,  we  shall  have  four  solutions  under  360°  in  all  cases  where  the  double 
angle  is  determined  by  a  single  function. 

The  logarithmic  solution  of  (319)  varies  with  the  signs  of  m  and  z.  Thus,  if  the 
equation  is 

sin  (a  -f-  z)  sin  z  =  m 
m  being  essentially  positive,  we  have  by  (133) 

cos1  £  a  —  cos2  (z  +  i  «)  =  sin  (a  -f-  2)  sin  z  =  m 

cos*  (z  +  i  a)  =  cos*  £  a  —  m 
and  by  (133)  again  this  is  solved  by 

cos*  <j>  =  m,  cos2  (z  -f  J  o)  =  sin  (^  -f-  J  a)  sin  (</> —  }  «) 

and  the  other  cases  are  solved  by  similar  methods. 


88 


PLANE  TRIGONOMETRY. 


172.  The  preceding  examples  will  suffice  to  indicate  the  method  to  be  followed  with 
all  the  equations  of  the  following  table.  The  solutions  of  the  equations  involving 
cosines  may  be  obtained  from  those  involving  sines,  by  exchanging  z  for  90°  ±  z,  or 
o  for  90°  ±  a. 

Logarithmic  solutions  of  the  first  four  will  be  obtained  by  imitating  the  process  of 
Art.  171. 


EQUATIONS. 


SOLUTIONS. 


1.  sin  (a  db  z)  sin  z  =  m 

2.  cos  (a  ±  z)  cos  z  =  m 

3.  sin  (a  rb  z)  cos  z  =  m 

4.  cos  (a  rb  z)  sin  z  =  m 

5.  sin  (o  ±  t)  =  m  sin  z 

6.  cos  (a  rfc  z)  =  m  cos  z 

7.  sin  (a  ;fc  z)  =  TO  cos  z 

8.  cos  (a  ±  z)  =  m  sin  z 

9.  tan  (a  ±  z)  tan  a  =  m 

10.  tan  (a  rb  z)  =  m  tan  z 


cos  (o  ±  2  z)  =  2  m  —  cos  a 

sin  (a  ±  2  z)  =  2  TO  —  sin  a 

sin  (a  ±  2  z)  =  sin  a  ±  2  m 

tan  ^  =  TO,  tan  (z  ±  £  o)  =  cot  (^  =F  45°)  tan  }  a 
tan  <j>  —  m,  tan  ( J  a  ±  z)  =  tan  (45°  —  <j>)  cot  $  o 
tan  ^  =  m, 

tan  (45°  —  J  a  =F  z)  =  tan  (45°  —  <t>)  tan  (45°  +  £  a) 
tan  0  =  TO, 

tan  (45°  —  |  a  =p  z)  =  tan  (45°  =F  0)  tan  (£  a  —  45°) 
tan  ^  =  m,  cos  (a  ±  2  z)  =  tan  (45°  =F  ?)  cos  a 

tan  $  =  TO,  sin  (2  z  ±  o)  =  cot  (<j>  =F  45°)  sin  o 


In  the  numerical  solutions  the  signs  of  the  angles  and  their  functions  must  be  care- 
fully observed.  The  signs  of  the  functions  should  be  prefixed  to  their  logarithms, 
according  to  Art.  99. 

The  auxiliary  angle  (j>  may  be  taken  numerically  less  than  90°  in  all  cases,  but  posi- 
tive or  negative  according  to  the  sign  of  its  tangent.  It  can  easily  be  shown  that  we 
shall  thus  obtain  the  same  values  of  z  as  by  taking  <j>  in  the  2d  quadrant  when  its 
tangent  is  negative,  or  in  the  3d  quadrant  when  its  tangent  is  positive. 


EXAMPLE. 

Find  z  from  (317)  when  a  =  65°  and  m  =  1-5196154.     By  (318) 

log  tan  <t>  =  log  n 


+  0'1817337 
56°  39'  9" 
-11°  39'  9* 
-  9'3143426 
-\-  9-6259483 


log  tan  (45°  —  <f>)  cos  a 

a-f  2z 


2z 

2 


45°  -0 
log  tan  (45°  —  0 

log  cos  a 
log  cos  (a  -f  2  z)  —  8-9402909 
95°  or  265°  or  455°  or  625° 
65° 

30°  or  200°   or  390°  or    560° 
15°  or   100°   or  195°   or  280° 


*  It  must  be  remembered  that  in  this  employment  of  the  signs  +  and  — ,  these 
signs  belong  to  the  natural  numbers  ;  and  when  the  logs,  are  added  or  subtracted,  the 
sign  of  the  result  is  to  be  determined  according  to  the  rules  of  multiplication  and 
division  in  algebra. 


SOLUTION  OF  TRIGONOMETRIC  EQUATIONS.  89 

173.  To  find  z  from  the  equation 

sin  (a  -f-  z)  =  m  sin  (ft  -}-  2). 

Put  z'  =  ft  -f-  z,  a'  =  a  —  /?,  then  this  equation  becomes 
sin  (a'  -\-  z')  =  m  sin  tf 

which  is  of  the  form  (309)  and  may  be  solved  by  (309*)  or  (311) ;  then  z  =  z'  —  ft. 

In  the  same  manner  equations  of  this  form,  involving  cosines  or  tangents,  may  be 
reduced  to  those  of  the  preceding  table. 

174.     To  find  k  and  z  from  the  equations 

k  sin  z  =  m  '• 


We  have,  by  division, 


V        (320) 
k  cos  z  =  n 


m 

tan  z  =  — 


which  gives  two  values  of  z,  one  less,  the  other  greater  than  180°; 
whence,  also,  two  values  of  k  from  either  of  the  equations 


7  _      m  n 

k  —  ~       — 


sin  2      cos  z 

The  solution  becomes  entirely  determinate  (z  not  exceeding  360°) 
as  follows : 

1st.  When  the  sign  of  k  is  given.  For  if  k  is  positive,  sin  z  has 
the  sign  of  m,  and  cos  z  the  sign  of  ??,  and  z  must  be  taken  in  the 
quadrant  denoted  by  these  signs.  If  k  is  negative,  the  signs  of  sin  z 
and  cos  z  are  the  opposite  to  those  of  m  and  n,  and  z  must  be  taken 
accordingly. 

2d.  When  z  is  restricted  by  either  the  condition  z  <  180°,  or 
z  >  180°.  For  under  either  of  these  conditions  the  tangent  gives 
but  one  solution.  If  z  <  180°,  k  has  the  sign  of  m  ;  and  if  z  >  180° 
k  has  the  opposite  sign  to  that  of  m. 

3d.  When  z  is  restricted  to  acute  values,  positive  or  negative.  For 
under  this  condition  a  positive  tangent  will  give  z  between  0°  and 
-f  90°  ;  and  a  negative  tangent,  between  0°  and  —  90°  ;  and  k  will 
always  have  the  sign  of  n. 

It  follows  that  ??i  and  n  being  any  given  numbers  whatever,  we 
may  always  satisfy  the  conditions  expressed  by  (320),  1st,  by  a  posi- 
tive number  k  and  an  angle  z  between  0°  and  360°  ;  2d,  by  a  num- 
ber k  (unrestricted  as  to  sign)  and  an  angle  z  <  180° ;  3d,  by  a 
number  k  (unrestricted  as  to  sign)  and  an  angle  z  >  180°  ;  and  4th, 
by  a  number  k,  and  an  angle  z  in  the  1st  or  4th  quadrant. 

12  H  2 


90  PLANE  TRIGONOMETRY. 

EXAMPLE. 

To  find  k  and  z  from  (320),  (k  being  a  positive  number),  when 
m  =  -  0-3076258,  n  =  +  0.4278735. 

k  sin  z  -  0-3076258 

k  cos  z  +  0-4278735 

(a)                                                   log  k  sin  z  -  9-4880228 

(6)                                                  logjfe  cos  z  +  9-6313147 

(a)  — (6)                                            log  tan  2  -9-8567081 

3  324°  17'  6"-6 

(c)                                                        log  sin  z  -  9-7662280 

(a)  — (c)                                                   log/;  +  9-7217948 

k  +  0-5269808 

Upon  this  problem  and  the  deductions  we  have  made  from  it,  rests 
the  method  of  introducing  the  auxiliary  angles  required  in  solving 
many  of  the  formulae  of  spherical  trigonometry.  It  is  applicable  to 
any  equation  that  can  be  reduced  to  the  form  of  that  solved  in  the 
following  article. 

175.   To  solve  the  equation 

m  cos  z  -|-  n  sin  z  =  q  (321) 

m,  ?i  and  q  being  given. 

The  first  member  will  be  reduced  to  the  form  k  sin  (<f>  -\-  z)  by  as- 
suming k  and  <p  such  that 

k  sin  <p  =  m,  k  cos  <p  =  n  (322) 

whence 

k  sin  <p  cos  z  +  k  cos  <p  sin  z  =  q 

sin  (<p  +  z)  =  |  (323) 

/C 

Therefore,  if  k  and  ^>  be  found  from  (322)  by  the  preceding  article, 
(k  being  limited  to  positive  values),  we  can  then  find  by  (323)  the 
value  <f>  ~H  z  and  therefore  of  z.  There  will  be  two  solutions  from 
the  two  values  of  <p  +  z  given  by  (323). 


SOLUTION  OF  TRIGONOMETRIC  EQUATIONS. 


If  we  restrict  <p  to  values  less  than  180°,  (as  we  may  do  according 
to  the  last  article),  we  may  find  it  by  the  equation 


m 

tan  tp  =  - 
n 


and  then  sin  (tp  -f  z)  =  -"  sin  tp  =  -*•  cos  <p 

m  n 

and  in  this  form  it  will  be  unnecessary  to  find  &.* 


(324) 


EXAMPLE. 


To  find  z  from  (321)  when  m  = 
and  q  =  —  04316893. 

By  (322)  and  (323). 

log  m  =  log  k  sin  <p  —  0*0211203 

log  n  =  log  k  cos  <p  +  9-8731402 

log  tan  <p—  0-1479801 

tp  305°  25'  20" 

log  sin  <p  —  9-9111059 

log  k  +  0-1100144 

log  q  —  9-6351713 

log  sin  (<p  +  z)  —  9-5251569 

,       f         199°  34'  40" 

!  t    or  340°  25'  20" 

(      -  105°  50'  40" 

!  \     or   35°    0'    0" 


-  1-0498332,  n  =  +  0-7466898, 

By  (324). 

log  m  — 0-0211203 
log  n  +  9-8731402 
log  tan  <p  —  0-1479801 
tp  125°  25'  20" 
log  sin  <p  -f-  9-9111059 
log  q  —  9-6351713 
ar  co  log  m  —  9-9788797 
log  sin  (f  -f  z)  +  9-5251569 
,        f         19°  34' 40" 
'  \  or  160°  25'  20" 
105°  50'  40" 
or  35°    0'    0" 


To  avoid  the  negative  value  of  2,  in  the  first  of  these  solutions,  we 
may  take  for  the  first  value  of 


tp  +  z,  360°  +  199°  34'  40"  =  559°  34'  40" 


The 


whence     z  =  559°  34'  40"    -  305°  25'  20"  =  254°    9'  20". 
second  solution  gives  a  like  result. 

If  we  suppose  tp  in  (324)  to  be  limited  to  acute  values  positive  or 
negative,  we  take  <f>  =  -  54°  34' 40",  which  gives  tp  +  z=  199°  34'  40", 
or  340°  25'  20",  whence  the  same  values  of  z  as  before. 

We  may  repeat  the  latter  part  of  the  work  with  cos  tp  for  verifi- 
cation. 


*  The  solution   is,  by    (323),  impossible  when  -3  is  greater   than   unity ;    and  by 

adding  the  squares  of  (322),  &  —  m2  -f-  n*  J  therefore  the  solution  is  impossible  when 
9*  >  m"  -f  ?i*. 


92  PLANE  TRIGONOMETRY. 

176.  To  solve  the  equation 

a  sin  (a  -f  z)  -f  b  sin  (/?  -f  2)  -f  csin  (y  -f  2)  -f  etc.  =  g.  (325) 

Developing  by  (36)  and  putting 

a  sin  o  -\-  b  sin  /3  -f"  c  sin  7  +  etc.  =  w» 
a  cos  o  -j-  b  cos  p  -\-  c  cos  y  -f-  etc.  =  n 

this  becomes 

m  cos  z  +  »  sin  a  •=  j 

which  is  solved  in  the  preceding  article.     The  same  process  applies  if  any  or  all  of  the 
terms  contain  cosines. 

177.  To  find  k  and  z  from  the  equations 

k  sin  (a  -f-  2)  =  m  ) 

Asin  (/»  +  .)=«  }    (326) 

The  sum  and  difference  of  these  equations  are,  by  (105)  and  (106), 


2  k  sin  [J  (a  +  0)  -f  z]  cos  J  (a  —  0)  =  m  -f  n 
2/fccos[$  (a  +  /j)+s]  sin  J  (a  —  /J)=m  —  n 
whence 


2  A  sin  „, 

cos  J  (a  —  /3) 


sin  i    a  — 


(327) 


from  which  2  &  and  J  (a  -(-  /?)  +  z  are  determined  by  Art.  174.     The  logs,  of  the 
second  members  of  these  equations  should  be  computed  separately,  for  the  purpose  of 
readily  discovering  the  signs  of  the  sine  and  cosine  in  the  first  members.     The  solution 
is  determinate  (according  to  Art.  174)  when  the  sign  of  k  is  given. 
From  (327)  we  find,  by  division, 


tan  *(<*-/*)  (328) 


which  requires  a  less  number  of  logs,  than  the  separate  computation  of  (327),  but  we 
are  obliged  to  refer  to  (327)  to  determine  (by  an  inspection  of  the  second  members) 
the  signs  of  the  sine  and  cosine. 
If  we  assume 


tan0  =  ^ 
m 


(329) 
re  may  compute  (328)  by  the  formula 

tan  [J  (a  -f  /?)  -f  2]  =  tan  (45°  +  <t>,  tan  J  (a  —  /?) 


SOLUTION  OF  TRIGONOMETRIC  EQUATIONS.  93 

EXAMPLE. 

In  (326)  given  a  =  200°,  P  —  140°,  TO  —  —  0'42345  and  n~  —  0*20123,  to  find  * 
and  k,  k  being  positive. 

By  (327). 

m  -f  n  —  0-62468 

m  —  n  —  0'22222 

J(«  +  0)  170° 

i(«-/3)  30° 

log  (m  +  n)  -  9-7956576 

log  cos  J  (a  —  ft)  -f  9-9375306 

(a)  log  2  k  sin  [i  (a  -f  ft)  -f  z]  —  9'8581270 

log  (m  —  n)  —  9-3467831 

log  sin  $  (a  —  19)  -f  9'6989700 

(b)  log  2  A  cos  [$  (a  +  |3)  +  z]  —  9'647813l 
(a)  —  (b)                                                     log  tan  [*  (a  +  ft)  +  z]  +  0'2103140 

i  (a  +  /3  +  z)  238°21'38"-6 

z  68°21'38"-6 

(c)  log  sin  [£  (a  +  ft)  +  a]  -  9'9301171 
(a)  —  (c)                                                                               log  2  i  +  9-9280099 

2  A  0-8472467 

k  0-4236234 

178.  A  more  general  solution  of  (326)  is  the  following.*    Let  y  be  any  angle  as- 
sumed at  pleasure,  and  in  (171)  let 


x  —  o      z,        y 
(distinguishing  the  e  of  (171)  by  an  accent)  ;  then  we  shall  find 
sin  (a  —  /?)  sin  (y  +  z)  =      sin  (a  —  y)  sin  (/?  +  «)  —  sin  (P  —  y)  sin  (a  -f  *) 
In  this  let  y  (whose  value  is  arbitrary)  be  exchanged  for  y  -f-  90°  ;  then 
sin  (a  —  P)  cos  (y  -f  z)  =  —  cos  (a  —  y)  sin  (P  -(-  z)  -f-  cos  (P  —  y)  sin  (a  -f-  2) 
Multiplying  these  equations  by  k  and  substituting  m  and  n  from  (326) 

A  sin  (o  —  /?)  sin  (y  +  z)  =  m  sin  (y  —  /3)  —  n  sin  (y  —  a)  j 

A  sin  (a  —  /?)  cos  (y  +  z)  =  rn  cos  (y  —  /3)  —  n  cos  (y  —  a)  J 

which  (y  being  assumed  at  pleasure),  determine  k  and  y  -f-  *• 
If  we  take  y  =  0,  we  find 

tan  z  —  —  m  sin  /^  +  n  sin  a 
m  cos  P  —  n  cos  a 

If  y  =  «, 

A  sin  (a  -(-  z)  =  m  -» 


If  y  =  /?,  we  have  a  similar  result. 

If  y  =  $  (a  -(-  j3)  we  obtain  the  solution  of  the  preceding  article. 

If  k  is  required,  without  first  finding  z,  we  have,  by  adding  the  squares  of  (330) 

k  sin  (a  —  ft)  =  y  [m*  -f  n'  —  2  m  n  cos  (o  —  /?)]  (332) 

*  GAUSS.     Theoria  Motus  Corporum  Oaelestium,  Art.  78. 


94  PLANE  TRIGONOMETRY. 

179.   To  find  k  and  z  from  the  equations 

k  cos  (a  -(-  z)  =•  m 
k  cos  (/3  -\-  z)  =  n 


(333) 


These  are  reduced  to  the  form  (326)  by  substituting  90°  +  a  and  90°  -f  /?  for  a  and  /?. 
We  find,  however,  by  a  process  similar  to  that  of  Art.  177, 


2  A  sin  [}  (a  +  /J)  +  «]  = 
2  A  cos  [i  (a  +  0)  +  «]  = 


n  —  TO 

sin  £  (a  —  /?) 

n  4-  TO 
cos  J  (a  —  /?) 


cot  [  J  (a  +  0)  -f  z]  =  tan  (45°  -f  <j>)  tan  $  (a  —  /?) 


(334) 


(335) 


EXAMPLE.  In  (333)  given  o  =  280°  16',  ft  =  200°  10',  m  —  —  G'62342,  and 
n  r=  0'69725,  find  2  and  k,  k  being  positive. 

Arts,  z  =  207°  5'  34"'4    A  =  1-0273643 

180.  The  more  general  solution  of  (333)  may  be  found  directly  from  (172),  but  it 
will  be  simpler  to  obtain  it  from  (330)  by  substituting  90°  +  a  for  o,  and  90°  -f  P 
for  /J,  whence 


k  sin  (o  —  /?)  sin  (y  -\-  z)  =  —  m  cos  (7  —  /?)  +  n  cos  (y  —  o) 
k  sin  (a  —  /?)  cos  (y  +  z)  =      m  si°  (y  —  /?)  — 7l  sin  (7  —  °) 


(336) 


y  being  arbitrary  as  before. 
If  y  =  0,  we  find 


tan  z 


—  m  cos  j$  -\- n  cos  a 

—  m  sin  /3  -f-  »  sin  a 


If  y  =  a, 


A  sin  (a  +  *)  = 


sin  (a 


COS  (tt  -f  2)  =  TO 


(337) 


If  y  =  $  (a  +  /?),  we  obtain  the  solution  (334). 

If  A  is  rajuired  directly,  the  sum  of  the  squares  of  (336)  gives 

k  sin  (a  —  /?)  =  y  [TO*  +  nj  —  2  m  n  cos  (a  —  /?)] 

as  in  Art.  178. 

181.  The  solutions  of  the  preceding  articles  may  be  applied  to  a  single  equation  of 
the  form 

n  sin  (a  -f-  z)  =  wi  sin  (ft  -)-  z) 

which  is  a  more  general  form  of  (309).    For  if  we  assume 

k  sin  (a  -\-  z)  —  m 
we  have  k  sin  (/?  +  z)  =  n 

whence  k  and  z  are  found  by  Arts.  177,  178. 


EQUATIONS  OF  THE  SECOND  AND  THIRD  DEGREES.          95 

182.  In  like  manner,  if  the  proposed  equation  is 

n  cos  (a  -j-  z)  =  m  cos  (/?  +  2) 
we  assume 

k  cos  (o  -f  z)  =  m 

whence  k  cos  (/?  -f  z)  =  n 

and  £  and  z  are  found  by  Arts.  179,  180.     As  the  sign  of  k  (in  this  and  the  preceding 
article)  may  be  arbitrarily  assumed,  there  will  be  two  solutions. 

NUMERICAL  EQUATIONS  OF  THE  SECOND  AND  THIRD  DEGREES. 

183.  To  solve  the  equation 

=  0  (338) 


when  q  is  essentially  positive,  and  p  either  positive  or  negative. 
We  have  from  (144),  exchanging  x  for  0, 

tan*  i  i>  —  2  cosec  0  tan  £  0  +  1  =  0  (339) 

and  (338)  may  be  reduced  to  this  form  by  substituting 


in  which  we  may  take  the  radical  only  with  the  positive  sign,  since  we  may  assume 
x  and  z  to  have  the  same  sign.     We  thus  reduce  (338)  to 


which  compared  with  (339)  gives 

—  2  cosec  <j>  —  -£—,  z  =  tan 


or  sin  <t>  —  —    J,  x  =  l/g~tan  J  £  (340) 

P 

which  gives  two  values  of  </>  less  than  360°  and  consequently  two  values  of  x.  If  6  be 
the  least  of  these  two  values  of  0  less  than  360°  (=  2  TT),  all  the  values  of  <t>  which 
have  the  same  sine  are 

6,  it  —  8,  2  TT  -f  0,  3  TT  —  0,  etc. 

and  all  the  values  of  tan  £  <t>  are 

tan  J  #,  cot  J  0,  tan  $  0,  cot  £  0,  etc. 

Hence  the  two  roots  of  (338)  are  found  by  the  formulae 

sin  6  =  —  *l£S,  Xi  —  i/-qtnn  %  ft,  x2  =  Vo^coi  f  0  (341) 

P 

in  which  0  may  be  always  taken  <  90°  with  the  sign  of  its  sine,  and  \/  q  is  to  be 
regarded  as  a  positive  quantity. 

As  long  as  2  j/  q  is  not  greater  than  p,  this  solution  is  possible,  but  when  2  \/  q  >  p, 
sin  0  is  not  possible,  and  both  roots  are  imaginary  ;  which  agrees  with  what  is  shown  in 
algebra. 


96  PLANE  TRIGONOMETRY. 

184.   To  solve  the  equation 


q  =  0  (342) 

when  —  q  is  essentially  negative,  p  being  eitf^er  positive  or  negative. 
We  have,  by  (143) 

tan*  J^  +  2cot0tanJ0  —  1  =  0 


and  (342)  is  reduced  to  this  form  by  substituting 


whence  z1  -j  —  —  z  —  1=0 

1/9 

The  required  solution  is  therefore 

2  cot  0  =  —  £—  >  z  =  tan 

1/9 


or  tan  <t>  =    J5,  z  =  V^tan  J  ^  V343) 

P 

If  0  is  the  least  value  of  0,  all  the  values  of  <f>  which  have  the  same  tangent  are 

6,  «•  -f  0,  2  TT  +  0,  3  TT  -f  0,  etc. 

and  all  the  values  of  tan  i  <j>  are 

tan  J  0,  —  cot  J  0,  tan  J  0,  —  cot  }  0,  etc. 

Therefore  the  two  roots  of  (342)  are  found  by  the  formulae 

at,  =  V/^tan  J  0,  xa  =  —  1/7  cot  $  0  (344) 

P 

in  which,  as  before,  the  radical  is  to  be  taken  as  positive,  and  0  <  90°  with  the  sign  of 
its  tangent. 

In  this  case  both  roots  are  real,  since  tan  0  is  always  possible. 

185.  To  solve  a  numerical  equation  of  the  third  degree.  It  is  shown  in  algebra  that  any 
equation  of  the  third  degree  may  be  reduced  to  one  in  which  the  2d  term  is  wanting  ; 
we  need  consider  therefore  only  the  form 


0  (345) 

To  resolve  this,  put 


we  find 

Now  z  may  be  decomposed  into  two  parts,  y  and  z,  in  an  infinite  variety  of  ways,  and 
we  may  therefore  suppose  that  y  and  z  are  such  as  to  satisfy  the  condition 


which  reduces  the  first  term  of  the  preceding  equation  to  0,  and  gives  the  two  con- 
ditions 


Put  y*  =  tlt    z*  =  ty  ;  then  we  have 


EQUATIONS  OF  THE  SECOND  AND  THIRD  DEGREES.          97 


so  that,  by  the  theory  of  equations,  t^  and  tt  are  the  two  roots  of  an  equation  of  the 
second  degree  in  which  the  absolute  term  is  —    ^ 

mi 

term  is  6  ;  that  is,  they  are  the  roots  of  the  equation 


second  degree  in  which  the  absolute  term  is  —    ^  and  the  coefficient  of  the  second 

mi 


If  then  we  find  the  two  roots  t,  and  tt  of  (m)  by  the  preceding  methods  we  shall 
have 

X  =  y  +  2=:1H  +  1H  (») 

It  will  be  necessary  to  consider  the  sign  of  a  in  the  equation  (m). 
1st.   When  a  is  positive,  (m)  comes  under  the  form  (342)  and  the  solution  by  (344) 
gives 

" 


and  by  (n) 

x  =  ^  *  (^  tan  J0  -  &  cot  i  0) 

and  if  we  assume 

tan  £  ^>  —  i/'  tan  $  0 
this  becomes,  by  (142) 

z  =  —  2  -y  -  cot  0 

Collecting  these  results  we  have,  for  the  solution  of  (345),  when  a  is  positive, 


,       tan  J4>  =  |^tan  J0 

_  •     (346) 

x  =  -  2  -^  I  cot  ^ 

in  which  the  radicals  -J  —•  and  -J  —  are  to  be  considered   positive,  and  0  is  to  be 
\  27          \   t> 

taken  <  90°  with  the  sign  of  the  tangent.     But  two  of  the  three  values  of  •j/'  tan  J  8 
being  imaginary,  the  given  equation  has  but  one  real  root.* 

*  If  r  represent  the  real  value  of  \/  tan  i  ^  a°d  a»  as  the  two  imaginary  roots  of 
unity,  the  real  value  of  x  is 


and  the  imaginary  values  are 


or  since  a1  as  =  1 


IS 


98  'PLANE  TRIGONOMETRY. 

2d.    When  a  is  negative  and  —  4  a3  <  27  6*.     Equation  (m)  becomes 


and  is  of  the  form  (338)  ;  its  roots  are  therefore  found  by  (341)  which  gives 


. 

27  6» 


=       ~        tan  *  0 


=       -        cot 


"»<     7   »  *  =      —       &  tan  *  0  +  ^  cot 

or  if  we  put,  as  before,  tau  J  $  —  -fr  tan  J  6,  the  solution  of  (345),  when  a  is  negative,  is 


sin  Y  —  —  — A/ —          tun  4  </>  r^?  •»*/  tin 

b\       27 


(347) 


which  gives  one  real  root,  (the  other  two  being  imaginary,  as  above),  when  sin  0  is 
possible,  i.  e.  when  —  4  a*  <  27  61.* 


Substituting  the  values  of  a,  and  a. 


and  also 
we  find 


r  =  tan  £  0  —  •=  cot  i  <j> 

r 

z,  ^=  ^ /  _£  (cot  0  -(-  coeec  ^  ;/  —  3) 
*    3 

x,  =  -» I  ^L  (cot  <t>  --  cosec  <f>  -j/  —  3) 
v   3 

or  finally,  a:,  being  the  real  root,  the  imaginary  roots  are 
TI  ~~        2  2 


*  The  two  imaginary  roots  will  be  found,  by  a  process  similar  to  that  employed  in 
the  preceding  note,  to  be 


xt=  -  - 

in  which  z,  is  the  real  root  found  by  (347). 


EQUATIONS  OF  THE  SECOND  AND  THIRD  DEGREES.          99 

3d.  When  a  is  negative  and  —  4  a3  >  27  6*.  In  this  case  sin  6,  in  (347),  is  impossible 
and  the  preceding  solution  fails.  This  is  the  irreducible  case  of  Cardan's  rule,  the  roots 
appearing  under  imaginary  forms,  although  it  is  known  that  they  are  all  three  reaL 
It  is,  however,  readily  solved  trigonometrically. 

In  Art.  77,  putting  <j>  for  x,  we  have 

sin*  <p  —  |  sin  0  +  J  sin  3  0  =  0  (»') 

and  (345)  may  be  reduced  to  this  form  by  substituting 

x  =  kz 

whence  ^"^Ti*  ~^~  7»  =  0  (n/) 

BO  that  we  must  have 


in  which  the  radical  is  to  be  taken  positive,  so  that  x  and  z  shall  have  the  same  sign. 
Comparing  (m')  and  (n')  we  have  also 


which  is  a  possible  sine  in  the  present  case.     We  may  therefore  take 

z  =  sin  <t> 
and  the  solution  is 


. 

sin  3  c*  —  li  A/  —  —  a;  =  2  sin  tf  A/  —  - 

^   '        a*  '3 

which  gives  three  real  roots  by  the  different  values  of  3  0,  which  have  the  same  sine. 
If  6  is  the  least  of  these  values,  all  the  values  of  3  <j>  are  expressed  by 

2  n  TT  -f  0    and     (2  n  -f  1)  n  —  6 
n  being  any  integer  or  0  ;  and  all  the  values  of  <t>  are  expressed  by 


3  3 

Now  all  integers  are  included  in  the  forms  3  m,  3  m  -f-  1  and  3m  —  1. 
If  n  =  3  m,  the  above  values  of  <f>  are 


whence 

sin  0  =  sin  £  0,        sin  <f>  =  sin  £  (T  —  0) 

If  n  =  3  m  +  1,  we  find  in  the  same  way 

sin  ^  =  sin  J  (ir  —  5),  sin  ^  =  J  * 

the  same  as  before. 

If  n  =  3  wi  —  1,  we  find  both  values  to  be 

sin  *  =  —  sin  J  (ir  -f-  5) 


100 


PLANE  TRIGONOMETRY. 


so  that  there  are  but  three  different  values  of  sin  <j>.     Substituting  these  in  (348),  the 
three  roots  of  (345),  when  a  is  negative  and  —  4  a1  >  27  62,  are  found  by 


—  6)=      2  -J  —  J-  sin  (60° 
3 


.     (349) 


in  which  0  <  90°  with  the  sign  of  its  sine,  and  the  radicals  are  taken  with  the  positive 
sign. 

EXAMPLES. 
1.  Solve  (345)  when  a  =  —  6101315,  b  =  —  5'766578.     We  find 


log    A  —      =  —  0-002651* 

0  £tl 

which  being  greater  than  any  log.  sine,  we  take  its  arithmetical  complement  and  pro- 
ceed by  (349).    Then 

log  sin  0  =  —  9-9974349 
0  =  —  83°46/44// 


55'  34"-7 
9-6705571 

60°  —  J0    87°  55'  34"-7 
9-9997155 

—  (60°+i  #)  —  32°  4'  2o/x-3 
—  9-7251024 

0-4551811 

0-4551811 

0-4551811 

0-1257382 
1-335790 

log  i,    0-4548966 
x,  =  2-850339 

log  x,  —  0-1802835 
*,  =  —  1-514549 

2.  Solve  (345)  when  a  =  —  7,  and  6  =  7. 

Ans.  xl  =  1-356896,  xt  =  1*692021,  z,  =  —  3'048917. 

3.  Solve  (345)  when  a  =  1'5,  and  6  =  —  45. 

Ans.  The  real  root  =  3'41 63975. 

It  may  be  observed  that  the  algebraic  sum  of  the  three  roots  is  always  zero,  in  con- 
sequence of  the  absence  of  the  term  in  x1  from  the  given  equation.  This  is  easily 
shown  from  (349)  where  there  are  three  real  roots,  and  from  the  forms  in  the  notes 
p.  98,  where  there  are  imaginary  roots.  This  principle  furnishes  a  simple  verification 
of  the  values  found  by  (349). 


*  The  sign  —  here  belongs  to  the  number  of  which  this  is  the  logarithm. 


CHAPTER    XI. 

DIFFERENCES  AND  DIFFERENTIALS  OF  THE  TRIGONOMETRIC 

FUNCTIONS. 

186.  IN  the  applications  of  trigonometry,  it  is  often  required  to 
compute  a  function  of  one  angle  from  that  of  an  angle  which  differs 
from  the  first  by  a  small  quantity.     In  such  cases  it  is  generally 
most   convenient   to   compute  the  difference  of   the  two   functions, 
which  may  be  applied  to  either  to  obtain  the  other. 

187.  To  find  the  increment  of  the  sine  or  cosine  of  an  angle  corre- 
sponding to  a  given  increment  of  the  angle. 

Let  the  angle  x  be  increased  by  Ax,  (this  notation  signifying  dif- 
ference, or  increment  of  #),  and  let  the  corresponding  difference  or 
increment  of  the  sine  be  expressed  by  J  sin  x  and  of  the  cosine  by 
J  cos  x ;  we  have,  by  this  notation, 

A  sin  x  =  sin  (x  -f-  A  x)  —  sin  x 
A  cos  x  =  cos  (a;  -f  A  x)  —  cos  x 
and  by  (106)  and  (108) 

A  sin  x  =       2  cos  (x  -)-  |  A  #)  sin  £  A  x  (350) 

A  cos  a;  =  —  2  sin  (x  -}-  £  A  or)  sin  ^  J#  (351) 

which  are  the  required  formulae. 

We  here  consider  the  difference  always  as  an  increment,  i.  e.  an 
increase,  and  give  it  the  positive  (algebraic)  sign ;  its  essential  sign 
may,  however,  be  negative,  and  it  will  then  be  in  fact  a  decrement. 
Thus,  in  (351)  the  second  member  will  be  negative  so  long  as  x  < 
180°,  and  therefore  the  increment  of  the  cosine  is  negative;  that  is, 
from  0°  to  180°  the  cosine  decreases  as  the  angle  increases.  In  like 
manner  A  sin  x  is  negative  when  x  >  90°,  and  <  270°. 

188.  To  find  the  increment  of  the  tangent  and  cotangent.     We  have 

A  tan  x  =  tan  (x  -f-  A  x)  —  tan  x 
A  cot  x  =  cot  (x  -f-  A  x)  —  cot  x 
and  by  (11 6)  and  (11 9) 

A  tan  x  =  - =  sec  (x  +  A  x)  sec  x  sin  A  x  (352) 

cos  (x  +  A  x)  cos  x 

S!  II    .-/  OC 

A  cot  x  =  —  -  =  —  cosec  (x  -f  J  x}  cosec  x  sin  J  x  (353) 

sin  (x-\-  Jx)  sin  x 

1 2  101 


102  PLANE  TRIGONOMETRY. 

189.  To  find  the  increment  of  the  secant  and  cosecant.    We  have 

A  sec  z  =     sec  (z  +  A  z)  —  sec  z 
A  cosec  z  =  cosec  (z  -(-  A  z)  —  cosec  z 
or  by  (130)  and  (132) 

A  sec  z  =     8in  \X.J^S — 5lJ3Ll_~J[  (354) 

cos  (z  -f-  A  z)  cos  z 

A  cosec  z  =  ~  2_go8  (z  -f  i  A  x)jin  J_Ax  (355) 

sin  (z  -f  A  z)  sin  z 

190.  To  find  the  increment*  of  the  squares  of  the  trigonometric  functions  corresponding  to  a 
given  increment  of  the  angle. 

We  Lave 

A  sin*  z  =  sin*  (z  +  A  z)  —  sin*  z 
=  cos*  z  —  cos*  (z  -f  A  z) 
whtn.e  by  (133) 

A  sin*  z  =  —  A  cos*  z  =  sin  (2  z  -f  A  z)  sin  A  x  (356) 

From  (115),  (116),  and  (119)  we  deduce 

tan*  z  -  tan*  y  =  sin  (x  +  ?)  sin  ('  ~  V) 
cos*  z  cos*  y 

cot*  z  —  cot*  y  =       sm  (x  ~r  y/jij?  ^  ~  y) 
sin*  z  sin*  y 

whence 

A  tan*  z  =  sin  (    x  +     *]JMI^_Z  /gg7  j 

cos*  (z  -f-  A  z)  cos7  z 


Acot*z  =  =  (358) 

sin1  (x  -}-  A  z)  sm*z 

From  (16)  we  have 

see*  (z  -f  A  z)  =  tan*  (z  +  A  z)  -f  1 

sec*  z  =  tan*  z  -f-  1 
the  difference  of  which  gives 

A  sec*  z  =  A  tan*  z  (359) 

and  in  the  same  manner,  from  (17), 

A  cosec1  z  =  A  cot*  z  (360) 

and  the  values  of  A  tan*  z,  A  cot*  z,  may  be  substituted  in  (359)  and  (360). 

191.  When  the  increment  of  an  angle,  or  arc,  is  infinitely  smalt, 
it  is  called  the  differential  of  the  angle,  or  arc ;  and  the  correspond- 
ing increments  of  the  trigonometric  functions  are  the  differentials  of 
these  functions. 

The  differential  of  x  is  denoted  by  dx  •  of  sin  x  by  d  sin  x,  etc. 


DIFFERENCES  AND  DIFFERENTIALS.  103 

192.  To  find  the  differentials  of  the  trigonometinc  functions  from  the 
differential  of  the  angle. 

Let  the  angle  x  and  its  increment  J  x  be  expressed  in  the  unit  of 
Art.  11  ;  or,  which  is  equivalent,  let  x  and  Jx  be  the  arcs  which 
measure  the  angle  and  its  increment  in  the  circle  whose  radius  =  1. 
It  is  evident  that  the  less  the  arc,  the  more  nearly  does  it  coincide 
with  its  sine  or  tangent;  therefore,  when  Ax  is  infinitely  small,  or 
becomes  dx, 

sin  dx  =  dx  sin  ^  dx  =  ^  dx 

This  may  be  demonstrated  more  rigorously  thus.  When  dx  is 
infinitely  small,  we  have  cos  dx  =  1,  whence 

sin  dx 

--  =  cos  ax  =  1 

tan  dx 

sin  dx  —  tan  dx 

but  the  arc  cannot  be  less  than  the  sine,  nor  greater  than  the  tan- 

gent, and  therefore 

dx  =  sin  dx  =  tan  dx 

Again,  when  Jx  is  infinitely  small,  or  becomes  dx,  we  must,  ac- 
cording to  the  principles  of  the  differential  calculus,  reject  it  when 
connected  with  finite  quantities  by  the  signs  +  or  —  ;  thus  we  must 
substitute  x  for  x  +  dx,  or  for  x  +  ^  dx. 

Upon  these  principles  we  find  the  differentials  directly  from  the 
finite  differences  (350),  (351),  (352),  (353),  (354)  and  (355)  as  follows  : 

d  sin  x  =  cos  x  dx  (361) 

d  cos  x  =  —  sin  x  dx  (362) 

d  tan  x  =  sec2  x  dx  —  (1  +  tan2  x)  dx  (363) 

d  cot  x  =  —  cosec2  x  dx  =  —  (1  -+-  cot*  x)  dx  (364) 

d  sec  x  =  tan  x  sec  x  dx  (365) 

d  cosec  x  =  —  cot  x  cosec  x  dx  (366) 

193.  In  the  same  manner  the  equations  (356),  (357),  (358),  (359)  and  (3*60)  give 

d  sin1  x  —  —  d  cos*  x  =  sin  2  x  dx  (367} 

d  sec*  x  =  5!5-?£(fa  (368) 

cos*  z 

(369) 


COS*  X  COS*  X 

d  cot1  x  =    d  cosec*  x  =  =^1°!*  dx  (370) 

* 


(371) 


104  PLANE  TRIGONOMETRY. 

194.  Although  the  equations  (361),  (362),  (363),  (364),  (365)  and 
(366),  are  rigorously  true  only  when  dx  is  infinitely  small,  they  may 
be  used  when  dx  is  a  finite  difference,  instead  of  the  equations,  (350), 
(351),  (352),  (353),  (354)  and  (355),  provided  dx  is  sufficiently  small 
to  be  considered  equal  to  its  sine  without  sensible  error,  and  is  also 
very  small  in  comparison  with  x.     This  is  very  frequently  the  case  in 
practice,  and  the  differential  equations  are  then  preferred  on  account 
of  their  simplicity.     It  is  only  necessary  to  observe  that  dx  must  be 
expressed  in  arc,  i.  e.  in  terms  of  the  unit  radius  ;  if  it  is  given  in 
seconds,  it  may  be  reduced  to  arc  by  Art.  9. 

195.  To  find  the  differential  of  an  angle  from  the  differentials  of  its 
functions. 

From  (361)  we  have 

(372) 


cos  a; 


but  it  is  convenient  in  this  case  to  employ  the  notation  of  inverse 
functions,  Art.  87.  Thus,  if  y  =  sin  x,  x  —  sin  ~*  y,  and  the  preced- 
ing equation  becomes 


In  the  same  manner  from  (362),  etc.,  we  find 

<*«*>-'?  =       ~d  (374) 


dtan-1^-^-  (375) 
1-rjT 

dcoi~ly=TTy>  (376) 

f-r-f-r  77^-  (377) 


d  cosec  l  y  =  -  (378) 


CHAPTER  XII. 

DIFFERENCES  AND  DIFFERENTIALS  OF  PLANE  TRIANGLES. 

196.  IN  trigonometrical  investigations  it  is  o, 
often  necessary  to   determine  the  effect  of  a 

small  change  in  one  of  the  data,  upon  the  com- 
puted parts.  Thus,  Fig.  29,  if  A,  AB  and 
A  C,  of  the  plane  triangle  A  B  C}  are  the  data,  A 
and  A  C  is  subject  to  an  error  of  C  Cf,  the  required  parts  will  be 
subject  to  errors  which  are  respectively,  the  differences  between 
A  CB  and  A  C'  B,  A  B  Cand  A  B  C',  B  C  and  B  C'.  In  the  same 
figure,  the  data  may  be  supposed  to  be  A,  A  B  and  AB  C,  and  the 
angle  ABC  may  be  regarded  as  subject  to  the  error  CB  C'  which 
produces  the  corresponding  errors  in  the  remaining  parts.  In  the 
same  manner,  the  data  may  be  A,  A  B,  and  ACS,  A  CB  being 
variable  ;  or,  A,  A  B,  and  B  C}  B  C  being  variable.  In  all  these 
instances,  A  and  A  B  are  constant,  while  the  remaining  four  parts  are 
variable,  and  may  be  considered  as  receiving,  simultaneously,  certain 
increments  which  are  related  to  each  other.  We  propose,  then,  to 
solve  the  general  problem  : 

In  a  plane  triangle,  any  two  parts  being  constant,  and  the  rest 
variable,  to  determine  the  relations  between  the  increments  of  the 
variable  parts. 

It  is  evident  that  the  solution  of  this  problem  resolves  itself  into 
an  investigation  of  the  differences  of  two  triangles  which  have  two  parts 
in  common.  We  shall  consider  the  several  cases  successively  ;  distin- 
guishing the  triangle  formed  from  the  given  one  by  the  application 
of  the  increments  as  the  derived  triangle. 

197.  CASE  I.  A  and  c  constant.   The  six  parts  of  the  given  triangle, 
ABC,  Fig.  29,  being  A,  B,  C,  a,  b,  c,  those  of  the  derived  triangle 
formed  by  varying  all  but  A  and  c,  are  Ay  B  -f  J  B,  C  +  J  (7, 
a  -f-  Ja,  b  -f  J6,  and  c.     In  these  two  triangles  we  have 

A  +  B+  O=  180° 
A  +  B  +  4B+  C+  JC=  180° 


whence  JJ5  +  JC=0,         JB  =  —  4C  (379) 

14  105 


106  PLANE  TRIGONOMETRY. 

Also  in  the  two  triangles  we  have 

a  =  c  sin  A  cosec  C  (m) 

a  -f-  Aa  =  c  sin  A  cosec  (C-f-  JC)  (n) 

the  half  difference  of  which  by  (355)  is 

i  j    _       c  sin  ^.  cos  ((7+  ^  -^ff)  sin  ^  JO  ,    , 

sinCsin(C'+  JC)  W 

,       , 


sin  |  AB  sin  |  J  C  sin  (C  +  AC) 

The  half  sum  of  (m)  and  (n)  by  (131)  is 


_1_  1    A      —  "  cos 

sin  Csin(<7+  JC) 
which  combined  with  (p)  gives 


tanJJC       tan(C+|JC) 
From  (260)  we  have 


~         c  sin 
tan  C= 


b  —  c  cos  A 
whence 

6  —  c  cos  A  =  c  sin  A  cot  C 
b  -\-  Ab  —  c  cos  A  =  c  sin  A  cot  (C-}-  AC) 

the  difference  of  which  by  (353)  is 

c  sin  A  sin  JC 


,       . 


sin  C  sin  (C  + 
therefore 

_^A_=__J^_  =        _£ (382) 

sin  AB          si"  ^      ~:-  "*  '    A™  V       ; 


DIFFERENCES  OF  PLANE  TRIANGLES.  107 

This  equation  gives  by  (135) 

\Ab  =  a 

sin  |  JO  cos  \AG~  sin  (O+  JO) 

and  dividing  (380)  by  this 

Aa  _  cos  (C+ It  AC)  ^^ 

Ab          cos  \  AC 

It  is  to  be  observed  that  the  increments  (or  half  increments)  of  the 
angles  must  be  deduced  from  their  sines  or  tangents,  since  it  is  only 
by  these  functions  that  a  small  angle  can  be  accurately  determined. 
Moreover,  a  small  arc  being  nearly  equal  to  its  sine  or  tangent,  the 
equations  (380),  (381)  and  (382)  express  very  nearly  the  ratios  of  the 
increments  of  the  sides  to  the  increments  of  the  angles,  or  rather  to 
those  increments  reduced  to  arc  by  Art.  9,  or  Art.  54. 

198.  CASE  II.  A  and  a  constant.  We  have  as  in  the  preceding 
case  AB  =  —  JO;  and  in  the  two  triangles 

b  sin  A  =  a  sin  B 
(b  +  Ab)  sin  A  =  a  sin  (B  +  J  B) 

the  difference  and  sum  of  which  give 

l  J  b  sin  A  =  a  cos  (B  +  $  A  B)  sin  £  J  B  (p) 

(b  +  |  Ab)  sin  A  =  a  sin  (B  +  \  A  B)  cos  \  AB 

whence  by  division 

4J6  4J6  64-4J6  tnc*A\ 

—a—      -  — 2 =  -         ^—  (384) 

tanlJi?  tan^JO      tan(£  +  £J£) 

In  the  same  way 

\AG^  = i  Jc      = c  +  ^Ac ,385 

tan  ^  JO  tan  %  A  B      tan  (O  +  £  J  O) 

From  the  equations 

c  sin  A  =  a  sin  O 
(c  -f  J  c)  sin  vl  =  a  sin  (O  +  JO) 
we  find  \  Ac,  sin  A  —  a  cos  (O  +  \  AC]  sin  ^  JO  ((/) 


108  PLANE  TRIGONOMETRY. 

which  combined  with  the  equation  (  p)  gives,  since  sin  £JO=  —  sin 


Jc  cos(0 

From  (p)  we  also  have 


sin     JO  sin  JS 


which,  when  J  6  is  to  be  found  from  J  B}  is  more  convenient  than 
(384).     In  the  same  way  from  (q) 


c  cos 


JO 


sin  £  JO  sin^J^  sin  C 

199.  CASE  III.     b  and  c  constant.     We  have 

c  sin  -B  —  6  sin  C 
c  sin  (B  +  J  J8)  =  6  sin  (O-f  ^C) 
the  sum  and  difference  of  which  give 

c  sin  (B  +  i  J£)  cos  ^  J5  =  6  sin  (C+  i  4C)  cos  ^  JO        (ja) 
c  cos  (B  +  |  J  B)  sin  |  J  5  =  6  cos  (C  +  £  JC)  sin  ^-  JO         (?) 
the  quotient  of  these  gives 


By  (224)  we  have 

a  =  b  cos  C-\-  c  cos  B 
a  +  J  a  =  &  cos  (O+  4  C)  +  c  cos  (B  + 

the  sum  and  difference  of  which  give 

a  -f  £  Ja  =  6  cos  (C-f  £  JO)  cos  £  JO+  c  cos  (5  -f  J  J£)  cos 

-^Ja  =  6sin(C+  £  JO)  sin  £  JO+  c  si 
These  expressions  are  reduced  by  (  p)  and  (q)  to 


and  by  division 


tan     AC 


DIFFERENCES  OF  PLANE  TRIANGLES.  109 

In  the  same  way  we  have 

|Ja       _  a  +  ^Ja  (m) 


Since  the  sum  of  the  three  angles  is  constant, 
AA  +  AB+  JO=0 


therefore  by  (115) 

sinA(JJ5+  JO) 

tan  4-  AB  +  tan  A  JO— ^-* 

cos  4  J.Bcos4  JO 

£i  £ 


sin 


Substituting  (£)  in  (r)  we  find 
sin  \  AC 


sin     JJ.  a-\-     Aa 


By  differencing  the  equation 

a2  =  b*  +  c2  —  2  be  cos 
we  find  instead  of  (392)  and  (393) 

\Aa      __  be  sin  (A  4- 
sinj  A  A  ~  a  +  \ 

K 


s^  AC 
which  substituted  in  (s)  gives 

j  Ja      _  csin(^  +  ^  AB) 
sin^  J^  "  cos^JC 

and  in  the  same  manner 


sin^JJS  6  cos  (0  +  4  JO) 

whence  also  (395) 

a-\-  %  A  a 


110  PLANE  TRIGONOMETRY. 

200.  CASE  IV.     A  and  B  constant.     We  have 


,       sn 

6  =  —     -  a 

sin  A 

,    ,    A  ,       sin  B  , 
b-\-  Ab  =  -     -(a- 
sm  J. 

whence  J  6  —  —     -  Aa 

sin  J. 

In  this  case  the  third  angle  is  also  constant  and  there  are  but  three 
variables  related  by  the  equation 


(397) 


sm  ^4.       sm  B      sin  (7 


This  case  is  not  strictly  included  in  the  general  problem  as  stated 
in  Art.  196,  since  the  two  triangles  have  not  two  parts  in  common. 

201.  The  second  members  of  the  equations  (380),  (381),  (382), 
(383),  (384),  (385),  (386),  (387),  (388),  (389),  (390),  (391),  (392), 
(393),  (394),  (395),  (396),  involve  the  increments  themselves,  which 
are  the  quantities  sought.  It  is  therefore  necessary,  in  many  cases, 
to  solve  these  equations  by  successive  approximations. 

For  a  first  approximation  we  consider  the  increments  in  the  second 
member  to  be  =  0,  employing  B  for  B  -f-  £  A  JS,  etc.,  and  taking 
cos  £  J  B  =  1,  etc.  This  will  evidently  produce  but  a  slight  error  so 
long  as  the  increments  are  small  as  compared  with  the  entire  parts 
of  the  triangle.  We  then  obtain  a  second  approximation,  by  recom- 
puting the  equation  in  its  complete  form,  employing  in  the  second 
members  the  approximate  values  of  the  increments.  With  these 
second  values  we  may,  in  the  same  way,  obtain  a  third  approxima- 
tion, etc.  Theoretically,  it  requires  an  infinite  number  of  such 
approximations  to  arrive  at  a  perfect  result  ;  but  in  practice,  the 
tenths  or  hundredths  of  seconds  being  the  limits  of  accuracy,  it  is 
rare  that  more  than  a  second  approximation  is  necessary. 

It  is  also  to  be  observed  that  in  computing  the  values  of  small 
quantities  such  as  the  increments  in  question,  we  may  employ  logar- 
ithms of  only  four  or  five  decimal  places  and  take  the  angles  to  the 
nearest  minute.  This  is  in  fact  one  of  the  chief  advantages  of  com- 
puting by  differential  formulae,  rather  than  by  the  direct  formulae 
applied  to  each  of  the  two  triangles  successively. 


DIFFERENTIAL  VARIATIONS  OF  PLANE  TRIANGLES.     Ill 


EXAMPLE. 
In  a  plane  triangle  whose  parts  are 


a  =  6053 


-B  =  35°ll'3"-4       C=86°7'7"-7 
b  =  4082  c  =  7068 


let  A  and  a  be  constant  while  b  is  diminished  by  50'5  ;  to  find  the 
change  in  the  angle  B. 

We  have  in  this  case  Jb  =  —  50'5  ;  and  by  (387) 


.     ,    4T>_         Y  A  b  sin  B 

~  b  cos  (B 


sin 


IST  APPROX. 

2o  APPROX. 

\Ab 

-  25-25 

b 

4082 

£ 

35°  11' 

35°  11' 

i-  AB 

0 

-15' 

B-\-  l  A  B 

35°  11' 

34°  56' 

log  \  A  b 

-  1-4023 

) 

ar.  co.  log.  b 
log  sin  B 
ar.  co.  1.  cos  (B  -\-  ^  A  B) 
log.  sin  ^  A  B 

6-3891 
9-7606 
0-0876 
-  7-6396 
-  15'  0" 

V              —  7-5520 

0-0863 
—  7-6383 
-  14'  56"-8 

It  is  evident  that  changing  the  angle  B  -f  i  ^  B  by  only  three  seconds 
would  not  affect  the  fourth  place  .of  its  cosine;  a  third  approximation 
is  therefore  unnecessary,  and  we  have  finally  J  J5  =  —  29'  53"*6. 
As  the  log.  sines  of  small  angles  do  not  vary  proportionally  with  the 
angles,  it  will  conduce  to  accuracy  to  employ  the  methods  explained 
in  Art.  115. 

DIFFERENTIAL  VARIATIONS  OF  PLANE  TRIANGLES. 

202.  The  equations  (380),  (381),  (382),  (383),  (384),  (385),  (386), 
(387),  (388),  (389),  (390),  (391),  (392),  (393),  (394),  (395),  (396)  and 
(397)  become  differential  by  making  the  increments  infinitely  small, 
that  is,  by  omitting  the  increments  when  connected  with  finite  qnan- 


112 


PLANE  TRIGONOMETRY. 


tities  by  the  signs  +  or  — ,  and  substituting  the  increment  itself  for 
its  sine  or  tangent,  and  unity  for  its  cosine,  (Art.  192.)  The  char- 
acter d  must  also  be  substituted  for  J.  These  changes  being  made, 
we  easily  deduce  the  following  differential  relations. 

CASE  I.  A  and  c  constant. 


da 


dB 


da 


dC      sin  C 

—  =  cosC 
db~C°8 


CASE  II.  A  and  a  constant. 


db^ 
dB 

cU 

dC 


dB  =  -dC 

db  ,         D 

= — -  =        6  cot  B 


dc 
dJB 

d± 
dc 


c  cot  C 

cos  B 
cos  C 


(398) 


(399) 


CASE  III.  b  and  c  constant. 


dB 
dC 


da 
dC 


tan  B 
tan  C 

da 
dB 


=  —  a  tan  C 


da 
~d~A 


—  c  sin  B  =  b  sin  C 


dC 
dA 


dB 
dA 


b        n 

-  cos  C 


a 


(400) 


DIFFERENTIAL  VARIATIONS  OF  PLANE  TRIANGLES.        113 

CASE  IV.  The  angles ,  At  B}  C,  constant, 
da  d  b  d  c 


sin  A       sin  B       sin  C 


(401) 


203.  These  differential  relations  are  often  employed  when  the  in- 
crements are  very  small,  instead  of  the  equations  of  finite  differences. 
We  have  already  seen  that  the  equation  of  differences  often  requires 
to  be  solved  by  successive  approximations,  the  first  approximation 
being  in  fact  obtained  by  employing  the  corresponding  differential 
equation.  In  all  cases  therefore  where  a  second  approximation  in  the 
use  of  finite  differences  could  not  alter  the  result  of  the  first,  it  is 
plain  that  the  differential  equation  is  sufficiently  accurate. 

The  increments  of  the  angles  must  generally  be  expressed  in  arc. 
Thus  if  dB  is  given  in  seconds  we  must  divide  it  by  R"=  206264"-8, 
or  substitute  dB  sin  I"  for  dB. 

dA 

But  in  such  fractions  as  — — >  this  substitution  is  evidently  unne- 

d  B 

cessary  provided  the  two  increments  are  always  expressed  in  the  same 
unit,  as  minutes,  seconds,  etc. 


EXAMPLE. 
In  a  plane  triangle  whose  parts  are 

A  =  58°  41'  48"-9        B  =  35°  11'  3"-4         (7=86°    V  7"-7 
a  =  6053  b  =  4082  c  =  7068 

suppose  b  and  c  to  be  constant  and  the  angle  A  to  receive  the  incre- 
ment dA  =  20"-6  ;  find  da  and  dC. 
From  (400)  we  have 

da  =  dA  sin  1"  c  sin  B 

,~      — dA  c  cos  B 

dC= • — 

a 

log  dA  1-3139  log  (—  dA)  —  1-3139 

log  sin  1"  4-6856  log  c       3'8493 

log  c  3-8493  log  cos  B      9-9124 

log  sin  B  9-7606  ar.  co.  log  a      6-2180 

log  da  9-6094  log  dC—  1-2936 

da  0-407  dC— 19"-7 

16  K  2 


114  PLANE  TRIGONOMETEY. 

204.  The  error  of  employing  the  differentials  in  any  case  may  be  determined  ap- 
proximately by  developing  the  equation  of  finite  differences  and  comparing  it  with  the 
corresponding  differential  equation.  We  shall  select  a  simple  example. 

We  have  from  (387)  and  its  corresponding  differential  equation  in  (399) 


sin  B 
A6  =  b  cot  B  A.B  sin  1" 

the  first  of  which  when  developed  gives 

i  A6  =  b  cot  B  sin  J  A£  —  2  b  sin  ^  +  *  -^  sin  J  A.B  sin  J  A£ 

sin  2? 

or  substituting       sin  £  AJ?  =  J  A.B  sin  \"  ',  sin  }  AJ3  =  J  A.B  sin  1",  and  also  J?  for 
B  -\-  \  AJB  in  the  second  term,  which  will  affect  so  small  a  term  but  slightly, 

A6  =  6  cot  B  AJ5  sin  1"  -  f  (A.B  sin  lx/)* 

m 

Comparing  this  with  the  differential  equation  above,  the  error  of  employing  the  latter 
is  approximately 


-  -      (A£  sin  1")* 

which  for  A  B  =  1°  is  —  '000015  b. 

It  appears  from  this  example  that  the  error  is  expressed  by  a  term  involving  the 
square  of  the  increment  ;  and  if  we  develop  all  the  equations  of  finite  differences  we 
shall  find  that  they  differ  from  the  corresponding  differential  equations  by  terms  in- 
volving the  squares  and  higher  powers  of  the  increment.  Hence,  employing  ilte  dif- 
ferentials instead  of  the  finite  differences  amounts  to  neglecting  the  terms  involving  the  squaret 
and  higher  powers  of  the  increments. 

205.  The  differential  relations  above  obtained  could  have  been  deduced  more  di- 
rectly from  the  formulae  of  plane  triangles  by  differentiation,  employing  the  values 
of  the  differentials  given  in  Art.  192.  Thus  in  CASE  I,  A  and  c  being  constant,  if  we 
differentiate  the  equation 

a  =  c  sin  A  cosec  C 
we  have  d  a  =  c  sin  A  d  cosec  C 

=  —  c  sin  A  cot  C  cosec  C  d  C 
=  —  a  cot  CdC 
as  in  (398). 

The  student  may  exercise  himself  by  deducing  the  other  relations  of  (398),  (399), 
and  (400)  in  a  similar  manner. 


CHAPTER    XIII. 

TRIGONOMETRIC  SERIES.     DEVELOPMENTS  OF  THE  FUNCTIONS  OF 
AN  ARC  IN  TERMS  OF  THE  ARC,  AND  RECIPROCALLY.* 

206.  THE  investigation  of  trigonometric  series  is  most  readily 
carried  on  with  the  aid  of  a  few  elementary  principles  of  the  Differ- 
ential Calculus.  All  that  will  be  required  here  will  be  no  more  than 
is  generally  given  in  the  first  chapter  of  a  treatise  on  that  subject, 
namely,  the  differentiation  of  simple  algebraic  functions,  and  Taylor's 
Theorem.  We  shall  employ  the  following  expression  of  this  theorem  : 

,,        .     .   d.fy      h   .   d?  .  fy       h?    .    d?  .  fy         A3    . 

h=  •-  •- 


in   which   fy   denotes   what  /  (y  -f-  A)    becomes    when   h  =  0    and 


-Lf^')  *>    etc.,  are  the  successive  differential  coefficients,  or  de- 

dy          df 

rivatives  of  fy. 

207.   To  develop  sin  x  and  cos  x  in  terms  of  x. 

We  shall  first  develop  sin  (y  -f  x)  and  cos  (y  +  x)  by  (402).     By 
(361)  and  (362),  if 

fy  =  sin  y 


I  U>*  f    U  '  '      O1U      '/ 

we  have  — •  =        cos  y 

dy 


.  /V  d  cos  v 

j  j  =         -T-2  =  —  sin  y 
dy2  rfy 

rf'./v  d  sin  t/ 

— •«  = — z  =  —  cos  y 

dy 


d*.fy  d  cos  y 

-—  =  —        — *  =        sin  y 
dy<  dy 


*The  leading  results  of  this  Chapter  being  of  very  general  utility  and  constant 
application  are  printed  in  the  larger  type,  but  as  they  are  not  referred  to  in  the  subse- 
quent large  print  of  this  work,  and  moreover  require  a  limited  acquaintance  with  the 
Differential  Calculus,  the  student  can  omit  them  at  the  first  perusal,  and  pass  directly 
to  Part  II. 

115 


116  PLANE  TRIGONOMETRY. 

so  that  the  values  of  the  coefficients  of  the  series  (402)  recur  in  the 
order  -f  sin  y,  +  cos  y,  —  sin  y,  —  cos  y,  and  therefore  /  (y  -f  #)  = 

sin  (y  -f  x)  —  sin  y  -f  cos  y  -  -  sin  y  •—  —  cos  y  —  —  -f  etc.      (403) 

1  \*  2t  i'&'o 

If  we  commence  with 

fy  =  COS  y 

the  coefficients  will  recur  in   the  order  -f  cos  y,  —  sin  y,  —  cos  y, 
-f  sin  y,  and  (402)  will  give 

X  X  3^ 

cos  (y  -f-  x)  =  cos  y  —  sin  y  -  —  cos  y  —  -  -f-  sin  y  --  f-  etc.         (404) 

1  \'  2>  l*.2'o 

If  now  we  put  y  =  0  in  (403)  and  (404),  sin  y  =  0,  cos  y  =  1,  the 
alternate  terms  of  the  series  vanish,  and  we  have 

-'+-+f         (405) 


cos  x  =  I  —  —  -f  —  —  ---  -"  --  h  etc.  (406) 

1-2      1-2-3-4      1-2-3-4-5-6 

It  may  be  observed  that  (406)  can  be  deduced  from  (405)  by  dif- 
ferentiation. 

208.  The  series  (405)  and  (406)  are  directly  available  for  the  con- 
struction of  the  trigonometric  table.  For  this  purpose  x  in  the  series 
must  be  expressed  in  arc,  since  (361)  and  (362),  upon  which  the  pre- 
ceding demonstration  rests,  require  #  to  be  in  arc,  Art.  9. 

EXAMPLE. 
Find  cos.  10°.     Reducing  10°  to  arc,  by  Art.  9,  we  have 

x  =  10  X  -01745329  =  -1745329 
and  computing  separately  the  positive  and  negative  terms  of  (406), 

--^2-  =  —  -01523086 
j^~  -  -00003866  _  _£_  =  _  -00000004 

1-00003866  -  -01523090 

—  -01523090 

cos  10°  =  -98480776 

agreeing  with  the  tables,  which  give  -9848078.     The  student  may, 
for  practice,  verify  any  other  sine  or  cosine  of  his  table. 


TRIGONOMETRIC  SERIES. 


117 


209.  To  develop  tan  z  in  terms  of  x. 

Representing  the  coefficients  in  the  series  (405)  and  (406)  by  letters,  we  have 


tan  x  = 


in  which 


1  —  a,  x2  -j-  a4  x*  —  a6  x6  +  etc. 
as= 


(407) 


etc. 


If  we  perform  the  division  of  the  numerator  by  the  denominator,  we  perceive  that 
the  result  will  be  a  series  containing  only  odd  powers  of  x,  and  commencing  with  the 
term  x.  But  as  the  law  for  the  successive  formation  of  the  coefficients  is  not  easily 
shown  in  this  way,  we  shall  resort  to  the  following  process.  Assume  the  series  to  be 


tan  x  =  «!  x  -f-  cs  Xs  -f-  c5  x5  -f-  etc. 

and  differentiate  it ;  we  find,  by  (363),  after  dividing  by  dx, 
I  +  tan*  x  =  Cj  -f  3  c3  x1  +  5  c5  x*  +  etc. 
or,  since  from  the  division  of  (407)  we  know  that  el  =  1, 

tan1  x  =  3  c,  x1  -f  5  cs  x4  +  7  c7  x8  -f  9  c9  a8  +  etc. 
The  square  of  (m)  is 


(m) 


tan1  x  =  ct  Cj  Xs 


x*  -f 


which  compared  with  (71)  gives 


x8  -|-  etc. 
-fete. 
+  etc. 
+  etc. 


cc 


etc. 


+  c6  a,  -f  c, 


etc. 


where  the  law  of  derivation  is  obvious.     We  have  preserved  the  factor  cv  although  it 
is  equal  to  unity,  in  order  to  render  this  law  more  apparent. 

Since  the  first  and  last  terms  of  these  expressions  are  equal,  as  also  the  terms  equally 
distant  from  them,  we  may  write  them  as  follows  : 


9 

'n=~(2 
etc. 


-  2c7, 
etc. 


c6) 


118  PLANE  TRIGONOMETRY. 

in  which  form  any  coefficient  «,„  +  l  when  n  is  even,  is  expressed  by  —  terms  all  of 

• 

whose  coefficients  are  =  2  ;  and  when  n  i$  odd,  by—   —  terms  all  of  whose  coefficients 

2 
are  2  except  the  last,  which  is  1.* 

If  we  now  substitute  the  value  of  q  =  1,  and  deduce  the  numerical  values  of  the 
coefficients  successively,  we  shall  find 


210.    To  develop  cot  2  in  terns  of  x. 
If  we  invert  (407)  we  have 


(40g) 
3-5  s-  2--'  ss--'      ^ 


(409) 


x  —  as  x3  +  «5  x5  —  etc. 

and  the  first  term  of  the  actual  division  is  —  ,  the  second  term  —  (a.  —  a.)  x.   and 

x 
the  succeeding  terms  evidently  involve  only  the  odd  powers  of  x.     Therefore  let 

cot  x  =  --  dtx  —  rfs  Xs  —  djX5  —  etc.  (o) 

x 

Tlie  coefficients  cannot  be  determined  by  the  method  of  the  preceding  article  in  con- 
sequence of  the  negative  exponent  in  the  first  term  ;  but  they  are  directly  deducible 
from  those  of  the  series  for  tan  x.  We  have  by  (142) 

tan  x  =  cot  x  —  2  cot  2  x  (  p) 

Now  the  series  (o)  being  true  for  any  value  of  x  will  give  cot  2  x  by  substituting  2  x  for 
x,  whence 

2  cot  2  x  =  —  —  2s  dl  x  —  2*  dt  x3  —  2«  dbx>  —  etc. 

x 

Subtracting  this  from  (o)  we  have  by  (p) 

tanz=(2l  —  1)  d,  x  +  (2*  —  1)  rf3xs  +  (2«  —  1) 
Designating  the  coefficients  of  (408)  by  cu  Cj,  c6,  etc.  we  have  also 
tan  x  =  ct  x    -f  ca  &    +  e6  z*    +  etc, 

and  the  comparison  of  these  two  values  of  tan  x  gives 


2*  — 1       (2.—  1)(2  +  1)      1-3 
2*  —  1  ~  (2*  —  I)  (2J+I)  ~~  3-5 


2«  —  1       (2»  —  1)(2»+1)      7'9 
etc.  etc. 


2"+  »  —  1 


*Euler,  and  after  him  Cagnoli  and  others,  make  these  coefficients  depend  upon 
those  of  the  series  sin  x  and  cos  x,  but  the  number  of  given  quantities  by  which  each 
coefficient  is  expressed  is  double  the  number  required  in  the  method  of  the  text. 


TRIGONOMETRIC  SERIES.  119 

Substituting  the  values  from  (408) 

e»=l  *  =  |  «  =  ^etc- 

and  reducing  the  coefficients  to  their  simplest  forms,  we  find  the  series  (o)  to  be 

I      *        a?         2x*  x7  2x»  ,.,Av 

cot  x  =  ------  —  —  --  -  -----  etc.  (  410) 

x       3       3»'5        35'5-7        3s'5'-7         s-'-- 


211.   By  a  process  similar  to  that  of  Art.  209,  but  which  we  leave  to  the  student, 
we  find 

,    ,    x3    ,    5x*    ,     61  3*     ,    277  x8 


,.„* 
.  (411) 


And  from  (408)  and  (410)  by  means  of  the  formula 

cosec  x  =  £  (cot  J  x  -f-  tan  J  x) 
we  find 

1    i    x  7  x3  31  x5  127  x' 


,  ,10v 


212.   To  develop  sin"1  y  in  terms  of  y.     (See  Art.  87). 
Let  x  =  sin"1  y  (or  sin  x  =  y)  ;  then  by  (373) 

dy 


Developing  the  second  member  by  the  Binomial  Theorem, 

dx  ,   .    1*3  1'3'5    6  t  \ 

^  =  i  +  iy!+^'+^r6/  +  *-  («) 

As  this  contains  only  even  powers  of  y,  the  series  from  which  it 
would  be  obtained  by  differentiation  must  contain  only  odd  powers 
of  y  ;  therefore,  let 

*  =  a\y  +  a^f  +  Oay5  +  «?/  + etc-  (n) 

There  will  be  no  term  independent  of  y  if  we  limit  x  to  values  be- 
tween 0  and  :£  90°,  for  then  when  y  =  0  we  must  also  have  x  =  0.* 
Differentiating,  we  have 

dx 

—  =  «!  +  3  ff^2  +  5  a,#4  +  7  a^y*  -f-  ete. 

«y 

which  compared  with  (ra)  gives 

1  1-3  1-3-5 

0^=1         3as  =  -         5a5  =  —         7  «7  =  _fi  ete. 

*  The  series  (413)  obtained  under  this  limitation  expresses  but  one  of  the  values  of 
sin"1  y,  but  if  we  denote  the  series  by  s,  we  shall  have  by  (95)  the  following  expres- 
sion, including  all  the  values, 

sin-1  y  =  n  TT  -(-  ( —  1  )n» 
n  being  an  integer,  positive  or  negative,  or  zero. 


120  PLANE  TRIGONOMETRY. 

therefore  (n)  becomes 

y3  ,    1-3    y5       1-3-5    y7  , 

,c+_.+_r+tlBi    (418) 


It  is  unnecessary  to  develop  cos"1  y  since  we  have 


i  -     i 

cos    y  =  --  sin     y 


213.  To  develop  tan"1  y.     Let  x  =  tan"1  y,  then  by  (375) 


from  which  we  infer,  as  in  the  preceding  problem,  that  the  required 
series  contains  only  odd  powers  of  y  ;  therefore  let 


x  =  aly-\-  atf  +  agf  +  o7y7  -f  etc.  (n) 

dx 

then  -  =  «i  -f  3  c^  -f  5  a^4  -|-  7  c^y6  +  etc. 

dy 

which,  compared  with  (m),  gives 

al  =  l          3  03  =  —  1          5a6  =  l          7  Oj  =  —  1  etc. 
so  that  the  series  is 

}y7-f  etc.  (414) 


214.  To  compute  the  ratio  (—  n)  of  the  circumference  of  a  circle  to 
its  diameter. 

We  have  heretofore  assumed  this  ratio  to  be  known  from  geometry, 
where  it  is  found  by  means  of  circumscribed  and  inscribed  polygons 
which  are  made  to  differ  from  the  circle  by  as  small  a  quantity  as  we 
please  ;  but  (414)  enables  us  to  express  its  value  in  a  series.  We 

have  tan  -  =  1,  therefore  if  we  make  y  =  1   in  (414)  we  have 

K  1  *.  .    II        »  1        I        1  SA1C\ 

-  =  l-3+---  +  --etc.  (415) 

But  this  series  converges  too  slowly  to  be  of  any  use.  To  obtain  a 
rapidly  converging  series  y  must  be  a  small  fraction.  We  might  em- 

ploy tan  -  =  --  -  (Art.  29),  but  in  consequence  of  the  radical,  it  is 
6         /3 


TRIGONOMETRIC  SERIES.  121 

simpler  to  resolve  '-  into  two  or  more  arcs  whose  tangents  are  known, 

and  to  compute  the  value  of  each  of  these  arcs  by  the  series.     To 
effect  this  let 


then  by  (123) 


whence 


-A  =  tan-1 1  +  tan"1  i'  (416) 

4 


7T  t  +  V 

tan  -  =  1  = 


4  1  —  ttr 

1  —  t 
l  +  t 


from  which,  assuming  any  value  of  i  at  pleasure,  the  corresponding 
value  of  tr  is  found. 

If  we  take  t  =  £,  we  find  t'  =\;  therefore  by  (416)  and  (414) 


-  —  tan"1  |  -f  tan"1 


A  few  terms  of  these  series  give 

-  =  -4636476  +  -3217506  =  -7853982 
4 

»  =  3-14169 

more  accurately      ic  =  S'14159  26535  89793 

1  3 

If  we  take  I  —  ":'    we    find  tr  =  -*    but  the  above  supposition  is 

evidently  the  best  adapted  for  rendering  both  series  sufficiently  con- 
vergent.* 

215.   To  resolve  sin  x  and  cos  x  into  factors. 

The  series  (405)  shows  that  x  is  a  factor  of  sin  x,  and  gives 

sinz  =  x(l  --  -  --  1  --  —  —  —  etc.)  (p) 

1-2-3  T  1-2-3-4-5 

*See  NOTE  at  the  end  of  this  chapter,  p.  124. 
18  L 


122  PLANE  TRIGONOMETRY. 

and  the  factors  of  the  series  within  the  parenthesis  must  evidently  be  of  the  form 

i-*  (.) 

A  being  a  constant,  but  having  a  different  value  in  each  factor.  The  required  factors 
must  be  such  as  to  reduce  the  second  member  of  (  p)  to  zero  whenever  the  first  member 
is  zero.  Now  sin  x  is  zero  for  the  value  x  =  0,  whence  x  is  a  factor  as  already  seen, 
and  also  for  x  =  ±  n  K,  n  being  any  integer  ;  therefore  the  general  value  of  (q)  is 

,-^», 

whence  A  =  r?  ir* 

which,  substituted  in  (7),  gives  as  the  general  factor 

i  __  *L 

nJ7T» 
Making  n  successively  =  1,  2,  3,  etc.,  the  equation  (  p)  becomes  therefore 

'"=•('  -£)('-£)  ('-£)•••  (419) 

The  factors  of  cos  x  in  (406)  must  also  be  of  the  form  (q)  ;  but  cos  x  is  zero  for 

x  =  ±  (2  n  +  1)-'  n  being  any  integer  or  zero,  and  the  general  value  of  (q)  is 
2t 

«       (2n+l)'7r' 

A.r 

(2n+  l)*7r* 
whence  A  =  '  ---  -r-^-i  - 

which,  substituted  in  (7),  gives  the  general  factor 


Making  n  successively  =  0,  1,  2,  3,  etc.,  we  have 

.  (420) 


216.  Logarithmic  sines  and  cosines.     By  means  of  (419)  and  (420)  the  logarithmic 
sines  and  cosines  of  the  tables  are  readily  computed. 

Put  z  —  TO  —  ,  then 
2 

—  i  - 


Mid  taking  the  logarithms 

log  .inY^k*!  +  1ogm 

h«  »          =  log    l  - 


TRIGONOMETRIC  SERIES.  123 

Developing  these  logs,  by  the  known  formula 

log  (1  —  n)  =  -  M  (n  -f  \  n2  +  \  n»  -f  etc.) 

(in  which  M=  modulus  of  common  logs.)  and  arranging  according  to  the  powers  of 
m,  we  have 

1  •        WIT  ,  7T       .      . 

log  sin  —  -  =  log  —  +  log  m 

—  1- 


—  -fte+i+i-H 

Jf  /  1     ,1,1         t   \ 

H  ---  h  etc.  I 
^  4sT66  / 


log  o» 


3 
—  etc. 

m»r  .   Jlf  /  1 

=   —  '- 


—  etc. 

By  summing  the  constant  numerical  series,  and  substituting  the  value  of  the  modulus 
M  =  '43429  44819  and  also  of  —  ,  these  formulae  become 

m 

log  sin  —  =  10-19611  98770  +  log  tn 

IB 


-m1  X  0-17859  64471 
-m4  X  0-01 4G8  89690 
-m«  X  0-00230  11796 
-m8  X  0-00042  58450 

-  m10  X  0-00008  49075 

log  cos  ^  =  10 

L 

-m*  X  0-5357893412 

-  m*  X  0-22033  45350 

-  m«  X  0-14497  43131 
-m8  X  0-10859  04688 

-  m'°  X  0-08686  0376G 


—  «WX  0-00001  76758 

—  mu  X  0-00000  37870 

-  m16  X  0-00000  08284 

-  m18  X  0-00000  01841 

—  etc.  (421) 


—  m"  X  0-07238  25502 

-  m1*  X  0-06204  20818 

-  m"  X  0-05428  68115 

-  MI"  X  0  04825  49426 

—  etc.  (422) 


In  these  expressions  10  is  added  to  render  the  logarithms  positive,  as  is  usual  in 
the  tables.* 


*See  the  preface  to  Callet's  Tables,  for  the  coefficients  of  these  series  carried  to  20 
decimal  places,  and  for  other  forms  given  them  by  which  they  are  rendered  still  more 
convenient. 


124  PLANE  TRIGONOMETRY. 

EXAMPLE. 
Compute  log  sin  9°.     We  have 

m  X  90°  =  9°  m  =  -1  log  ro  =  —  1 

and  therefore  by  (421) 

log  sin  9°  =  10-19611  98770  —  1- 

-  0-00178  59645 

—  O'OOOOO  14689 

—  O'OOOOO  00023 


=  10-19611  98770  —  1-00178  74357 
log  sin  9°  =    9-19433  24413 

217.  If  in  (419)  we  put  x  =  ~,  we  have 

2 


_  ff/2»  —  1\    /4*  —  1\    /§i^iJ\ 
2  1     21     /   \    4»     /    \    &     )' 

=  JL  (2-1)  (2+1)  (4-1)  (4  +  1)  (6-1)  (6  +  1)... 
2       2.        2.       4.       4.        6.        6.... 

whence  JL  =  A  .  A  .  A  .  ±  .  A  .  A  .  .  .  (423) 

2         133557 

which  is  Wattisfs  expression  of  IT. 


NOTE  to  page  121.  Computation  ofir.  Many  other  series  besides  those  of  Art.  214, 
may  be  given  for  computing  IT.  One  method  of  obtaining  them  is  to  resolve  tan"1  t 
and  tan"1  i'  into  two  others,  and  thus  make  \  n  to  depend  upon  three  or  more  arcs. 
From  (194)  we  easily  deduce 

tan"1  —  =  tan"1  —       -  +  tan"1  —  (a) 

m  m  +  n  mj  +  m  n  +  1 

tan-1  -1  =  tan-1  — tan-1 (b) 

m  m  —  n  m*  —  TO  n  +  1 

in  which  m  being  given,  n  may  be  assumed  at  pleasure.  The  numerators  of  the  frac- 
tions in  the  last  terms  will  reduce  to  unity  when  ma  +  1  is  divisible  by  n ;  if  therefore 
we  assume  n  and  p  so  as  to  satisfy  the  condition 

np  —  TO*  +  1  (e) 

we  shall  hare 

tan-1  —  =  tan-1  — (-  tan-1  — ^—  (a) 

TO  TO  +  71 


tan-1  —  =  tan-1  — tan"1  — —  (e) 

m  m  —  n  p  —  m 


TRIGONOMETRIC  SERIES.  125 

For  example,  let  m  =  3  ;   then  m1  -j-  1  =  10  =  1  X  10  =  2  X  5,  so  that  we  may 
take  n  =  1,  p  =  10  ;   or  n  —  2,  p  =  5,  whence  by  (d)  and  (c) 

tan-1  i  =  tan-1  -  +  tan"1  — 


=  tan-1  -  +  tan-1  - 
Substituting  in  (418) 

-  =  tan-1  -  +  tan-1  -  +  tan 

=  2  tan-1  -  —  tan-1  - 
2  7 

=  2  tan-1 1  +  tan-1  ]-  (/) 

=  tan-1 1  +  tan-1 1  +  tan-1 1  (s) 

The  equation  (/)  was  employed  by  CLAUSEN  of  Germany,  in  computing  n  to  200 
decimal  places,  and  (g)  was  employed  by  DASE,  also  of  Germany,  in  computing  n  to 
the  same  number  of  figures.  These  computations  were  carried  on  independently  of 
each  other,  and  the  results  when  communicated  to  SCHUMACHER,  (who  gives  them  in 
the  Astronomische  Nachrichten,  No.  589),  were  found  to  agree  to  the  last  figure. 
They  prove  the  value  previously  found  by  Mr.  Rutherford  to  be  erroneous  beyond  the 
150th  figure. 

By  means  of  the  formulas  (a),  (b),  (c),  (d)  and  (e)  we  may  again  subdivide  the  arcs 
as  often  as  we  please.  Thus,  it  is  easy  to  deduce 

-  =  2  tan-1  -  +  tan-'  -  +  2  tan-1  - 

=  2  tan-1  -  +  tan-1  -  +  tan"1  ^  -f-  tan"1  -J- 
.  4  tan-1 1  -  tan-1 1  +  tan-  1  +  tan-  ± 
=  4  tan— tan-1  — —  -j-  tan-1  — — 


which  last  is  known  as  Machin's  formula.     In  deducing  it  we  have  reduced  the  dif- 
ference of  two  arcs  to  a  single  arc  by  means  of  formula  (a). 

Another  method  is,  to  find  by  trial,  or  otherwise,  an  arc  a  multiple  of  which  is 

nearly  equal  to  —  >  and  whose  cotangent  is  a  whole  number;   and  then  deduce  the 

L  2 


126  PLANE  TRIGONOMETRY. 

difference  between  this  multiple  and  —     Thus  it  is  known  (from  the  trigonometric 

4 

tables)  that  cot  11°  15'  =  5  nearly ;  therefore  by  the  last  formula  of  Art.  79,  putting 

tan  x  =  — » 
5 


and  by  (194) 


therefore 


—  =  4  tan-1  -  —  tan-1  JL 
4  5  239 


as  was  found  above. 

If    we    resolve    tan-1  -  —  by  means    of    (c),    (d)   and   (e),  we    have    m  =  239, 

m1  -f-  1  =  57122  =  2-13*  =  n  p,  which   offers  several    suppositions  for  n  and   p;    if 
we  take  n  =  13»  =  169  and  p  =  2-13'  =  338,  we  find  by  («) 

i  =  4  tan-1  L  -tan-1 1  +  tan-  1 

which  was  employed  by  Rutherford. 
If  we  take  n  =  1,  p  =  57122,  we  find  by  (d) 

^=4tan-1|-tan-1-^-tan-1^. 


CHAPTER  XIV. 

EXPONENTIAL  FORMULAE.     TRINOMIAL  OR  QUADRATIC  FACTORS. 

218.   To  demonstrate  Euler's  formula 

cosx=%  (e  "/-1  +  e  -"-1)  (424) 

s\nx  =  —-     —  -(e"-1  —  e—  "-1)  (425) 

in  which  e  is  the  Naperian  base  of  logarithms,  or, 

e=1  +  I  +  F2  +  li3  +  etC- 
It  is  shown  in  the  theory  of  logarithms  that 

<-=1+(f)+w+ly+w+ete-      (426) 

where  for  brevity  we  write 

(1)  =  1  (2)  =  1-2  (3)  =  1-2-3,  etc. 

We  have  by  (405)  and  (406),  employing  the  above  notation, 

x3        x*        x6 
°"'=1-(l)  +  (4j-(6j+'te 

x         x3        x*        x7 


the  terms  of  which  are  the  same  as  those  of  (426),  but  with  alternate 
signs.  If  the  signs  in  these  two  series  were  all  positive,  the  sum  of 
the  two  would  be  equal  to  (426)  ;  and  it  is  evident  that  we  shall  make 
them  positive  by  substituting 

x*  =  —  #        or       x  =  z  y  —  1 
which  gives 

0031=1  +++  +etc- 


m 


128  PLANE  TRIGONOMETRY. 


--„„, 


whence 


But 


1  -1  x 

—  y  —  1,      «=-      -  =  —  x  y  —  \ 


1/--1          y  -1 

therefore 

cos  a?  —  y  —  1  sin  a;  —  e~xy'~1  (427) 

If  in  this  equation  we  substitute  —  x  for  x,  we  have,  by  (56), 

cos  x  -f  i/  —  1  sin  z  =  e'y~l  (428) 

The  sum  and  difference  of  these  equations  are 


(429) 
(430) 


whence  (424)  and  (425). 

219.  The  quotient  of  (430)  divided  by  (429)  is 


_  e 


220.  If  we  put 

y  =  e*y~l     =cosx-\~i/  —  1  sin  a;  (432) 

we  have  y~l  =  e~*  v~l  =  cos  x  —  y  —  1  sin  x  (433) 

and  (429)  and  (430)  become 

2  cos  *  =  y  +  y~l  (434) 

2  y  —  1  sin  x  =  y  —  y~l  (435) 

If  mx  be  substituted  for  x  in  these  formulae,  we  have 

ym=em*y~]     =  cosm#+  y  --  1  sin  ma;  (436) 

y~m  =  e~m*  y~l  =  cos  mx  —  y  —  1  sin  mx  (437) 

2  cos  mx  =  ym  +  y~m  (438) 

2  V  —  1  sin  mx  =  ym  —  y~m  (439) 


EXPONENTIAL  FORMULAE.  129 

221.  Moivre's   Formula.     The  value  of  ym  from  (432),  compared 
with  (436)  gives 

(cos  x  -f  y1  -  -  1  sin  x)m  —  cos  mx  -f  i/  —  1  sin  mx  (440) 

which  is  Moivre's  Formula.  It  shows  that  the  involution  of  the 
expression  cos  x  -{-  \/  •  -  \  sin  x  is  effected  by  the  multiplication  of 
the  angle. 

Again,  if  we  multiply  (432)  by 

cos  x'  +  ]/  -  -  1  sin  x'  —  e  »  Y-i 
we  have 

(cos  x  -\-  |/  -  -  1  sin  x)  (cos  xf  +  i/  —  1  sin  #')  =  e  (z  +  y)  l/~l 
—  cos  (x  -}-  x')  -}-  \/  -  -  1  sin  (a;  -f-  x'} 

which  shows  that  factors  of  this  form  are  multiplied  by  the  addition 
of  the  angles. 
We  have  also 

(cos  x-\-\/  —  1  sin  #)  (cos  x  —  y7—  1  sin  #)=cos2  x-\-sin2x=e°  =  1  (441) 
222.  Oeneral  form  of  Moivre's  Formula.     As  long  as  m  is  an  integer,  both  members 

of  (440)  can  have  but  one  value  ;  but  if  in  =  ^  the  first  member  becomes 

(1 

£ 

(  cos  x  -f-  i/  —  1  sin  x)*  =  i*'  (cos  x-\-  -\/  —  1  sin  x]f 

which  has  q  different  values*  in  consequence  of  the  radical  of  the  degree  q,  while 
the  second  member 

cos  ^  x  -f-  i/  —  •  1  sin  2-  a;  (a) 

1  9 

has  but  one  value. 

In  order  that  both  members  may  have  the  same  generality,  as  should  be  the  case 
with  every  analytical  expression,  it  is  necessary  to  suppose  that  we  take  for  the  arc 
x  not  merely  the  arc  less  than  the  circumference  which  has  the  given  sine  and  cosine, 
but  also  all  the  arcs  which  have  the  same  sine  and  cosine  ;  that  is,  c  denoting  the 
circumference,  all  the  arcs 

x,        x  -\-  e,        x  -f-  2  e,        x  -f  3  c,  etc. 

Now  there  is  an  infinite  number  of  these  arcs,  but  only  q  of  them  can  give  different 
values  to  (a)  ;  for  all  the  values  of  the  arc  in  (a)  will  be 


9  f        '   f 

P.  ,  +  !£_«,   £  x  +  [£iJLE2,  etc 


*  That  is  q  values  real  and  imaginary  ;  thus  it  is  shown  in  algebra  that  y  a*  =  -}-  a 
-  a  ;  «»  =  +  a,          a    ~  1  +  V  '  -3\          and         a    ~  1 


and  -  a  ;         &  «»  =  +  a,          a  (~  1  +  V  '  -3\          and         a  (~  1  ~  l^r_3\ 


V'  a4  =  -f-  a,  —  a,  -f-  a  y'  —  1,  —  a  ^x  —  1  ;  etc. 
17 


PLANE  TRIGONOMETRY. 

2  %  -}-  pc  has  the  same  sign  and  cosine  as  ^  z  ; 
99  9 

+  **-3  —  ^~£?  =  (£  x  -f-  ^  I  +  pc  the  same  sine  and  cosine  as  %  z  4-  ^,  etc., 
9  9  V  9/  99 

so  that  after  the  first  g  terms  of  the  above  series,  the  same  values  of  the  sine  and 
cosine  return  ad  infinitum.  Representing,  therefore,  the  circumference  by  2  nt  the 
equation  is  entirely  general  under  the  form 

P^ 
(cos  x  -\-  i/  —  1  sin  z)«  =  cos  2.  (2  w  TT  -f-  x)  -f-  j/  —  1  sin  ^  (2  n  TT  -(-  z)         (442) 

9  9 

in  which  n  is  any  number  of  the  series  0,  1,  2,  3  .    .    .    .9  —  1. 

223.   Trigonometric  expressions  of  the  real  and  imaginary  roots  'of  unity. 
If  z  =  0  in  (442)  it  gives 

£ 
(!)«  =  cos^  2n7r-j-1/  —  1  sin^  2mr  (443) 

9  9 

or  (l)m  —  cos  2  mmr  -4-  1/  —  Isin2win7r  (444) 

m  being  fractional  or  integral.     If  p  =  1,  (443)  gives 

^l^cosl^!i+1/_isin2jL^  (445^ 

9  9 

which  expresses  the  q  roots  of  unity  by  making  n  successively  0,  1,  2,  3  .    .    .  q  —  1. 
For  example,  let  q  =  4,  (  445  )  gives  for 

n  =  0,      i/'  1  =  cos  0  +  l/  —  1  sin  0  =  1 


n  =  2,      ^  1  =  cos  TT  -(-  |/  —  1  sin  TT  =  —  1 


as  found  in  algebra. 

If  x  —  -  in  (442),  it  gives 

Z 

(v/  _  1)-  =  cos  7ra(4n+])7r  +  T/  -  1  sin  m(4n  +  1)?  (446) 

2t  2t 

which  shows  that  an  imaginary  term  of  any  degree  can  be  reduced  to  a  binomial  of  the 
form  A  -\-  £  -\/  —  1. 

If  x  =  TT  in  (442)  we  find 

(—!)«  =  cos  m  (2ri  +  1)  TT  +  -\/  —  \  sin  711  (2  n  -f  1)  n  (447) 

224.    To   reduce   an   imaginary   quantity   of  the  form   (a  +  b  I/  —  l)m   to   ^'e  /orm 
4  +  .Bi/--l. 

Let  it  and  x  be  determined  from  the  equations 

k  cos  x  =  a,  k  sin  z  =  b 

by  Art.  174  ;  then,  by  Moivre's  Formula, 

(a  -f  6  i/  —  l)w  =  km  (cos  z  -4-  V'  —  1  sin  x)1* 

=  £m  (cos  mo-  -{-  |/  —  1  sin  mx) 
and  putting  A  =  km  cos  nuc,  B  =  km  sin  mx, 

(o  4-  6  T/  —  l)m  =  vl  +  JBl/-l. 


TRINOMIAL    OR  QUADRATIC   FACTORS.  131 


TRINOMIAL  OR  QUADRATIC  FACTORS. 

225.   To  find  the  quadratic  (trinomial)  factors  of  the  expression  ztm  —  2  zm  cos  </>  +  1  J 
m  being  integral. 

By  (438)  and  (434)  we  have 

ylm  —  2  ym  cos  mx  -f-  1  =  0 
y*    —  2  y    cos  x     +1=0 

Therefore  if  we  put  y  =  z,  mx  =  2  n  TT  -f-  0,  or  a;  =  —      t^,  we  have 

(448) 
(449) 

771 

As  these   two  equations  exist  at  the   same  time,  they  have  common  roots,  and  the 
second  is  therefore  a  divisor  or  factor  of  the  first;  but  this  factor  has  m  values  in 

consequence    of   the    m    values    of    cos   —  —     f    (Art.    222),  found    by   making 

m 

n  =  0,  1,  2,  3  .  .  .  m  —  1.     Therefore  the  m  quadratic  factors  of  (448)  are  all  ex- 
pressed by  (449),  and  we  have 


—  2Z"1  cos 


i>  -f  1  =  (z2  —  2  z  cos  4-  +  l] 
\  ml 


x   ... 

X  (z2  -  2  »  cos  2("-l)ff  +  ^  +  A  (450) 

V  in  I 

226.  To  obtain  the  simple  factors  of  (448),  we  have  only  to  find  the  two  simple 
factors  of  each  of  the  quadratic  factors  in  (450),  or  to  find  the  two  factors  of  the 
general  quadratic  (449).  Now,  by  the  theory  of  equations,  if  2X  and  zt  are  the  two 
roots  of  (449),  the  first  member  is  equal  to 

(»-«,)  (•-%) 

but  we  have  by  (432) 

.    . 
z  =  y  =  cos  x  4-  T/  —  1  sin  x  =  cos 


m 


which  gives  the  two  values  of  z  by  the  double  sign  belonging  to  |/  —  1.     Therefore 
the  simple  factors  of  (448)  are  all  included  in  the  form 

(451) 


132  PLANE  TRIGONOMETRY. 


EXAMPLES. 

1.  Find  the  quadratic  and  simple  factors  of 

z*  —  2  z2  +  1 

Here  m  —  2,  2  cos  0  =  2,  cos  #  =  1,  <t>  —  0  ;  and  by  (450), 

z*  —  2  z2  +  1  =  (  z*  —  2  z  cos  0  +  1  )  (z-  —  2  z  cos  *  +  1  ) 
=  (z2  —  2z-fl) 


by  (461) 

=  0  —  (cosO  +  v/  —  Isin  0)] 


X  [z  —  (cosO  —  |/  —  1  sin  0)] 
X  [z  —  (cos  T  +  i/  —  1  sin  ?r)] 
X  [«  —  (cos  TT  —  i/  —  1  sin  ?r)] 
=  (*-!)  (*-!)(«  +  !)(*+!) 
2.  Find  the  factors  of  z*  -f  2  2'  -f  1.     Here  m  =  2,  2  cos  <t>  =  —  2,  *  =  it,  and 


3.  Find  the  factors  of  2*  —  z1  +  1. 

«*  —  *'  +  1  =  (,"  —  2  z  cos  30°  +  1)  (z2  +  2  z  cos  30°  +  1  ) 


4.  Find  the  factors  of  z«  —  2  2s  +  1. 

=  (2'  —  2z+  1)  (ZJ  +2 


227.   To  /nrf  the  quadratic  factors  of  zm  —  1  when  m  is  odd. 
In  (450)  let  <t>  =  0,  it  becomes 


2jr   , 
m 

X  (#  —  2  z  cos  —  -f 
\  m 

x  .  .  . 

n  / i\  _  v 

(452) 


Now  m  being  odd,  m  —  1  is  even,  and  the  number  of  trinomial  factors  in  (452) 
exclusive  of  (z  —  I)1,  is  even  ;  but 


so  that  the  first  and  last  of  these  factors  are  equal.     In  the  same  manner  it  is  shown 
that  any  two  of   these  factors   equally  distant   from  the   first  and  last  are   equal  ; 


TRINOMIAL  OR  QUADRATIC  FACTORS.  133 

therefore,  uniting  these  equal  factors  and  extracting  the  square  root  of  both  members, 
we  have,  when  m  is  odd, 

a»_  1  =  (z  —  1)  X  (z2  —  2  z  cos  —  + 
\  m 


(453) 
/ 

228.  To  find  the  quadratic  factors  o/z"*  —  -  1,  when  m  is  even. 

When  TO  is  even,  m  —  1  is  odd,  the  number  of  factors  in  (452),  exclusive  of  («  —  I)1, 
is  odd,  and  the  middle  factor  will  not  combine  with  any  other.     This  factor  is  the 

and  contains 


[—  ) 


cos  -     —  =  cos  JT  =  —  1 
m 


and  is  therefore  equal  to 


so  that  uniting  the  remaining  factors,  and  extracting  the  square  root,  we  have,  when 
m  is  even, 

a"  —  1  =  (z  —  1)  (z  +  1)  X  (z2  —  2  z  cos  ~ 

V  TO 

X(*J  —  2zcos  — 

\  TO 

X  ---- 

(464) 

m 

229.   To  find  the  factors  of  zm  -f-  1,  when  m  is  odd. 
In  (450)  let  <t>  =  TT,  it  gives 


(a*  +  I)4  =  fza  —  2  z  cos  —  +  l 

\  TO 


-- 

TO 


and  it  is  easily  shown,  as  in  the  preceding  articles,  that  the  factors  equally  distant  from 
the  first  and  last  are  equal,  and  that  the  middle  term  is  z1  -{-  2  z  +  1  =  (z  -f  I)1 

M 


134  PLANE  TRIGONOMETRY. 

Hence  we  find,  when  m  is  odd, 

2»  +  1  =  (z  +  1)  X  (z2  —  2  z  cos  —  + 
V  m 


X    #  —  2  z  cos        + 
\  m 

x.... 

X  (2'  -  2  *  cos  (™^         +  1  (465) 


230.   To  /ncf  the  factors  of  zm  -\  -  1,  w/ien  m  ts  even. 
The  same  process  gives 


z»  +  1  =  (2»  —  2  z  cos  —  +  l\ 
\  m          I 

X  («*  —  2  z  cos  —  +  l) 

V  7/1  / 

x.... 

(456) 


231.  The  simple   factors  of  (453)  and   (454)  are  obtained   from   (451)   by  putting 
0  =  0,  and  those  of  (455)  and  (456)  by  putting  <j>  =  IT.     There  will  be  found  pairs  of 
equal  factors  as  in  the  preceding  articles,  but  all  the  different  simple  factors  will   be 
found  by  taking  only  the  positive  sign  of  the  radical  i/  —  1. 

232.  Any  function  of  the  form  2*"'  —  2  p  z™  -j-  q  may  also  be  resolved  into  quadratic 
factors.     It  is  only  necessary  to  reduce  it  to  one  of  the  preceding  forms.     By  resolving 
the  equation 

23m  _  2  p  z*  -f  q  =  0  (457) 

we  shall  find  from  its  two  values  of  zm 


and  if  we  put  the  absolute  term  in  one  of  these  factors  =  dc  am  (according  to  its  sign) 
it  becomes 


2»  ±  a™  =  am  f—  db  1\  =  a"  (z"»  ±  1) 

\am         / 


in  which  z  =  az',  and  the  factors  of  this  last  expression  may  be  found  by  one  of  the 
preceding  articles. 

If,  however,  the  values  of  z"1  in  (457)  are  imaginary,  i.  e.  if  p*  <  q,  this  method 
fails  to  discover  the  real  quadratic  factors,  and  we  must  proceed  as  follows.  Put 
q  =  o2m,  then  the  proposed  function  becomes 


in  which  z  =  oz'  ;  and  since  in  the  present  case  p  <  am,  -2-  is  a  proper  fraction,  and 

am 

we  may  put  -2-  =  cos  0,  which  reduces  the  given  function  to  the  form  (450). 
a* 


CHAPTER  XV. 


TKIGONOMETKIC  SERIES  CONTINUED.     MULTIPLE  ANGLES. 

233.  THE   true   developments   of   sin   mx   and   cos  mx  in   series,  when   m  is   not 
restricted  to  integral  values,  were  first  obtained  by  Poinsot,  and  form  the  subject  of  a 
memoir  read  by  him  before  the  French  Academy  of  Sciences,  in  1823.*    The  fol- 
lowing problem  is  the  basis  of  these  investigations. 

234.  To  develop  (k  -f-  V"  k?  —  l)m,  in  a  series  of  ascending  powers  of  k.     Let 

z  =  (k  -f  V '  K1  — 1)m  (a) 

and  assume 

Differentiating  (a)  and  putting 

/  =  ok 
dk 
we  find 


z'  =  m  (k 

the  square  of  which  gives 

m»  z 
Differentiating  this  and  putting 


X 


_fc \_          ma 

P_i)j-V(*_ 


dk 


we  find,  after  dividing  by  z', 

m*  z  —  ltzf  —  (A2  —  l)z//P  =  0 
Again,  differentiating  (6)  twice,  we  find, 

Substituting  in  (d)  the  values  of  z,  zx,  z",  given  by  (b)  and  (e),  we  have 


(d) 


=  mt  An 


.2.4, 


2.3  ^3 


A  -f-  m2  J.2 


+  mtAn 

—  nAn 

—  (n  —  1)  n  An 


(n 


in  which  each  of  the  coefficients  of  the  powers  of  k  must  be  zero.  To  discover  the 
law  which  governs  these  coefficients,  it  will  suffice  to  examine  that  of  the  general 
term,  or  the  coefficient  of  &",  which  is 

whence 

*       _* 

An 


(n  +  1)  (n  +  2) 


*  See  the  published  memoir,  "  Recherches  sur  F  Analyse  des  Sections  Angulaires," 
Paris,  1825. 

135 


136  PLANE  TRIGONOMETRY. 

•o  that  from  the  first  coefficient,  A0,  we  find  by  making  n  =  0,  2,  4,  6,  etc., 

A  -        m*   A 

*•*"    15* 

A-        m'~*A  -"V"'-*)    * 
3'4~At~       1-2-3-4       A° 

A  —  -»'  —  **       ___  m(m'  —   » 


5-6  1-2-3  -4-5  -6 

etc. 

and  from  the  second  coefficient,  Av  we  find  by  making  n  =  1,  3,  5,  etc., 


2-3 

,   _         »'- 
" 


2-3-4-5 

m'  —  5*  .  (OT»  —  1')  (m«  —  8«)  (ma  —  6') 

^~       ~6y~A"  2" 

etc. 

Therefore,  if  we  put 


the  equation  (6)  becomes 

2  = 

and  it  only  remains  to  find  A0  and  ^4^     In  (a),  (b),  (c)  and  (e),  put  i  =  0  ;  we  find 

»=(,/  —  1)»  =  4,  2'  =  m(1/-l)m-1  =  A 

Therefore  we  have,  finally, 

/  -  1)"-1  m  JT'  (458) 


235.  To  develop  (VI  —  A*  +  A  -j/  —  l)m  in  a  series  of  ascending  powers  of  h.     We 
have 


hv  —  l}m  =  (v  —  l)m  (h  +  V^f^l)m 
therefore  by  (468),  exchanging  k  for  h, 

(  vT=^»  +  h  v  -  \T  =  (v  —  !)m  \.(v  —  i)m  R+  (v  —  i)"-1  mff/] 

in  which  H  and  IP  are  what  K  and  .K7  become  when  h  is  put  for  k.    Combining 
the  imaginary  factors  in  the  second  member,  observing  that 

(V  ~  1)"  X  (V  ~  l)m  =  (V  !)m  =  0)' 


MULTIPLE  ANGLES.  137 

(which  must  not  be  put  equal  to  unity,  since  m  may  be  a  fraction,  and  unity  has 
imaginary  roots,)  and  also  that 

(v  -  !)m  x  (v  -  1)"-1  =  v  -  1  (  v  -  1)—1  x  (v  -  1)"-1  =  v-  1  (i)  ^ 

we  have 


tf+  A  y  -  1)™  =  (1)*  H+v—l  (1)  »  mH'  (459) 

in  which 


- 

236.  To  develop  the  sine  and  cosine  of  the  multiple  angle  in  a  series  of  ascending  powers 
of  the  cosine  of  the  simple  angle. 

When  m  is  an  integer,   this  problem  requires  us  simply  to  develop  sin  mx  and 

cos  mx  in  a  series  of  powers  of  cos  x  ;  but  when  m  is  a  fraction  =  JL,  the  angle  mx 

9 

has  q  values  which  have  the  same  sine  and  cosine,  (Art.  222),  if  we  consider  x  to  repre- 
sent all  the  angles  which  have  the  same  sine  and  cosine  as  the  simple  angle.  We  shall 
therefore  employ  Moivre's  Formula  in  its  general  form  (442),  or 

(cos  x  -f  V  —  1  §in  z)m  =  eaim  (2  n«r  4*  *)  4*  V*  —  1  sin  m  (2  n  n  -f-  x) 
Putting  k  =  cos  x  we  have  by  (458)  and  (446), 
(cos  z--f  V  —  1  sin  x)m  =  (k  +  V'P  —  l)m 
=  (T/  -  1)»  JT+  (v'  —  I)"-1  m  X"' 


.    /( 
—  I  sin  I  v 


^ 
mJt/ 


Comparing  the  real  and  imaginary  terms  of  these  two  values  of  (cos  x  -f  i/  —  1  sin  z)m, 
we  have 


sin  m  (2  n  TT  +  x)  =  sin  ^  ^  +  1)  ^     ^  +  g-n  ^(m-  1)  (4^  -f  1) 

If  m  is  a  fraction  =  -^,  each  member  of  these  equations  receives  q  values  by  taking 

? 

successively  for  n,  or  n7,  the  numbers  of  the  series  0,  1,  2,  3,  ...  q  —  1  ;  but  we  are  now 
to  show  what  values  of  n  and  n'  correspond  to  each  other  in  the  two  members.     Let 

x  =  -,  then  k  =  0,  K=l,  K'  —  0,  and  we  have 


18  M  2 


138  PLANE  TRIGONOMETRY. 

therefore   these  two  angles  can  only  differ   by  some  multiple  of  2  T,   or  we   must 
have 

m  (4n  4-  1)  TT TO  (4  n' -{- 1)  TT       „     .. 

~2~  T- 

whence  m  (n  —  n')  =  n" 

hut  TO  being  a  fraction  ?,  and  n,  nf  numbers  of  the  series  0,  1,  2,  .    .    .  q  —  1,  we  can- 
not have  m  (n  —  nx)  equal  to  an  integer  n",  unless  it  is  zero  ;*  therefore 

n  —  n'  =  0,  n  =  n' 

and  the  above  developments  are 

coswi  (2n7r  +  x)  =  cos  /j»(4»  +  l)ff\  .  _g-+  cos  ((m  ~  1U4n  +  !)  ff\  .  mjfi:         (460) 


sinm(2n7r  +  z)=:sin(-^"or  *>     j.^+sin  (v""~  V  ^ft  "^  ^  "  )  .  mJT/       (461) 
in  which 


.. 

1*2  1*2°3'4 

K>  =  cos  x  -  ?^1'  cos*  ,  +  ("'-  *;)  (™'  -^  cos*  x  -  etc. 

It  hence  appears  that,  in  general,  it  requires  the  combination  of  two  series  to  express 
the  cosine  and  sine  of  a  multiple  angle  in  powers  of  the  cosine  of  the  simple  angle, 
when  m  is  fractional. 

237.  When  m  is  an  integer,  one  of  the  terms  of  (460)  and  (461)  will  always  become 
zero,  and  we  shall  have  but  a  single  series  to  express  the  function  of  the  multiple  angle. 
The  first  members  become  in  all  cases 

cos  (2  mn  it  +  mx)  =-  cos  mx 
sin  (2  mn  TT  -(-  mx)  =  sin  mx 

and  the  second  members  vary  according  to  the  form  of  m.     In  (460),  if 

m  =  4  m',  cos  mx  =  K 

m  =  4  m/  -f-  1,  cos  mx  =  mKf 

m  =  4  m'  +  2,  cos  mx  =  —  K 

m  =  4  m'  -f-  3,  cos  mx  =  —  mK/ 

and  since  when  m  is  even,  the  series  K  terminates,  and  when  m  is  odd,  the  series  K' 
terminates,  these  four  equations  are  all  finite  expressions,  and  will  give  the  equations 
of  Art.  76,  by  making  m  =  1,  2,  3,  etc. 

In  (461),  if 

m  =  4  m',  sin  mx  =  —  mK 

m  =  4  m/  -(-  1,  sin  mx  =  K 

m  =  4  m'  -{-  2,  sin  mx  •=  mK' 

m  =  4  mf  -\-  3,  sin  mx  =  —  K 

*  Since  ^  is  supposed  to  be  reduced  to  its  lowest  terms,  p  and  q  are  prime  to  each 
q 

other;  therefore,  if  £J_n—    —1  is  not  zero,  q  must  divide  n  —  n' \  which  is  impossible, 

? 
since  the  greatest  value  of  either  n  or  n'  is  q  —  1. 


(462) 


MULTIPLE  ANGLES.  139 

In  these  formulae,  however,  the  series  do  not  terminate,  but  by  differentiating  (462) 
we  find  for 

mt 2* 

m  =  4  m',  sin  mx  =  —  m  sin  x  (cos  x cos1  x  -f-  etc.) 


(463) 


m  =  4  m'  -J-  1,        sin  mx  =  sin  x  (1  —         -  cos*  x  -j-  etc.) 

1*2 

m  =  4  m'  -{-  2,        sin  ma;  =  m  sin  a;  (cos  z  —  ~~~^  —  cos*  z  ~f~  etc>) 

m9 
mt  _  Jl 

m  =  4  m'  -f-  3,        sin  mx  =  —  sin  a;  (1  --  —  —  cos*  *  -j-  etc.) 

i  £i 

all  of  which  terminate  and  give  the  equations  of  Art.  75. 

238.   To  develop  the  sine  and  cosine  of  the  multiple  angle  in  a  series  of  ascending  povxra 
of  the  sine  of  the  simple  angle. 

We  take  as  before 

(cos  x  +  i/  —  1  sin  x)n  =  cos  m  (2  n  TT  -f-  x)  -f-  1/  —  1  sin  m  (2  n  ?r  -f  x) 
Putting  A  =  sin  x,  we  have,  by  (459)  and  (444), 
(cos  x  -f-  i/  —  1  sin  x)m  =  (V  1  —  A1  -+-  AT/  —  l)m 


=  (1)     ff  -f  |/  —  1  (l 
=  cos  mn'  TT  .  H  '  -(-  j/  —  1  sin  m  n'  K  .  H 
+  I/  —  ^  cos  (TO  —  1)  n'  TT  .  mU'  —  sin  (wi  —  1)  n'  ir  .  m  J 
Comparing  the  real  and  imaginary  terms  of  these  equations, 

cos  tn(2n7r-|-x)  =  cos  m  nf  IT  .  .ff  —  sin  (m  —  1  )  nf  if  .  m  Hf 
sin  m  (2  n  JT  -f-  a;)  =  sin  m  n7  TT  .  £T  +  c08  (TO  —  1)  nx  TT  .  m  IT' 


and  to  find  what  values  of  n  and  nf  correspond,  let  x  =  0,  then  h  =  sin  x  =  0,  JJ=  1, 
H'  =  0,  and  we  have 


sin  2  m  n  TT  =  sin  m  n'  IT 
from  which  we  infer  that  2  m  n  TT  =  m  nr  ir,  or  2  n  =  n',  and  hence 

cos  t»  (2  »  T  -f-  z)  =  cos  2  mn  TT  .  H  —  sin  2  (m  —  1)  n  IT  .  mH'          (464) 
sin  m  (2  n  ?r  -j-  x)  =  sin  2  mn  n  .  H  -\-  cos  2  (m  —  1)  n  JT  .  mJiF          (465) 

in  which  m  being  a  fraction  =  E,  n  is  any  number  of  the  series  0,  1,  2,  3,  ...  g  —  1  ; 
and  1 

l-t    sin'x-etc. 

'a        tt)    sin'x-etc. 


2-3  2-3-4-5 

239.  When  mis  an  integer,  the  first  members  of  (464)  and  (465)  become  cos  mx  and 
sin  mx  ;  and  the  coefficients  of  the  second  members  contain  only  multiples  of  2  IT  ; 
therefore  we  have 

cos  mx  =  H  sin  mx  =  mH' 

But  the  series  H  terminates  only  when  m  is  even,  and  the  series  H'  only  when  m  is 


140  PLANE  TRIGONOMETRY. 

odd,  and  we  must  also  employ  the  derivatives  of  these  equations  to  obtain  finite  ex- 
pressions in  all  cases ;  thus  we  have  also 

sin  mx  = —  cos  mx  — 


mdx  dx 

Therefore  differentiating  the  series  H  and  H/,  we  shall  have,  when 

m  =  2  m',  cos  mx.  =  1  —  —  sin1  x  -f-  etc. 

I'm 

m  =  2  m'  +  1>      cos  mx  =  cos  x  (1  —  m  ~ —  sin* x  +  etc. ) 

1  4 

m  =  2  mf,  sin  mx  =  m  cos  z  (sin  x  —  m  '          sin*  x  -f  etc.) 

4  O 
S  -]J 

TO  =  2  m'  +  1,       sin  mx  =  m  (sin  x sin5  x  -\-  etc.) 

2*3 


(466) 


(467) 


all  of  which  terminate,  and  give  the  equations  of  Arts.  77  and  78. 

240.  To  develop  the  sine  and  cosine  of  the  multiple  angle  in  a  series  of  ascending  powers 
of  the  tangent  of  the  simple  angle. 

We  have 

cos  m  (2  n  n  +  x)  -f  i/  —  1  sin  m  (2  n  IT  -f  x)  =  (cos  a;  +  v  —  1  sin  *)" 

=  cosm  x  (1  -f-  V  —  1  tan  x)* 
Expanding  by  the  Binomial  Theorem,  and  putting 


tan'*  +  etc. 
1*4*0 

we  have 

cosm  (2nT  +  x)  +  i/  —  1  shim  (2  n  ir  -|-  x)  =  cos1"  x  (T+|/  —  1  Tx) 

But  the  imaginary  and  real  quantities  are  not  yet  distinctly  separated  in  the  second 
member,  for  m  being  fractional  cosm  x  has  a  number  of  imaginary  values.  If  we  desig- 
nate its  real  value  by  cosm  x,  all  its  values  are  included  in  the  expression 

cos*  x  (l)m  =  cosm  x  (cos  2  mn'  IT  -f  y  —  1  sin  2  mn'  n) 
which,  substituted  above  for  cos"*  x  gives 

cos  m  (2  n  IT  -f  x)  -f  y  —  1  sin  m  (2  WT  -f  x)  =  cos"  x  (co82mnx  n  .  T  —  sin  2mnx  ir  .  2") 

+  1/  —  1  cosm  x  (sin  2  mn/  ir  .  T  -f-  cos  2  mm'  n  .  T') 

Comparing  the  real  and  imaginary  terms,  we  now  have 

cosm  (2n7r  +  x)=cos">x  (cos  2mn'ir  .  T—  sin  2  mn'  ir  .  T') 
sinm(2n7r  +  x)=co8"x  (sin2mn/7r  .  T-f  cos  2mn'ir.T') 

»nd  it  is  shown  as  in  the  preceding  problems  that  n  =  n',  whence 

cosm  (2n7r4-x)=cosmx  (cos  2  win*  .  T—  sin  2mn7r  .  T')  (468) 

sinm  (2  n  TT  -(-  x)  =  coem  x  (sin  2mn?r  .  T-f  cos  2mnir  .  2")  (469) 


MULTIPLE  ANGLES.  141 

in  which  m  being  a  fraction  =  %,  n  is  any  number  of  the  series,  0,  1,  2,  .    .    .  q  —  1  ; 


__ 
and  cosmz  denotes  only  the  real  value  of  ^(cos  xp. 

241.  By  the  division  of  (469)  by  (468) 

x       tan  2mn7r.  T+  Tf  IAnK. 

tan  m  (2  n  ir  -f-  x)  =  —  (470) 

T—  tan  2mmr.T' 

242.  When  m  in  an  integer,  both  the  series  T  and  Tf  terminate,  and  in  all  cases 
cos  2  mn  rr  =  1,  sin  2  mn  TT  =  0  ;  and  (468),  (469)  and  (470)  give 

cos  mx  =  cosmx  .  T  (471) 

sin  mx  =  cosm  x  .  Tf  (472) 

tan  mx  =  ~  (473) 

which  last  expression  embraces  all  the  equations  of  Art.  79.* 

243.  Before  the  memoir  of  Poinsot,  developments  were  given  for  the  multiple  arcs 
in  series  of  descending  powers  of  the  sine  or  cosine  of  the  simple  arc  ;  but  he  has  shown 
that  these  developments  are  impossible,  except  when  m  is  integral,  and  in  this  case  the 
series  are  the  same  as  the  preceding,  with  the  terms  written  in  inverse  order. 

244.  To  develop  any  power  of  the  cosine  of  the  simple  angle  in  a  series  of  sines  or  cosines 
of  the  multiple  angles,  the  cosine  of  the  simple  angle  being  positive. 

If  y  =  cos  x  +  |/  —  1  sin  x,     we  have,  by  (434)  and  the  Binomial  Theorem, 

(2  cos  z)m  =  (y  +  y-1)"  =  ym  +  mym-2  -f  m(m~1I  ym-4  _j_  eta 

% 

and  by  Moivre's  Formula, 

y*»  =  cos  m  (2  n  tr  -\-  x)  -\-  i/  —  1  sin  m  (2  n  TT  -j-  x) 
mym-t  —  m  egg  (m  —  2)  (2  n  K  -j-  x)  -f-  wi  i/  —  1  sin  (m  —  2)  (2  n  IT  -(-  x) 


' — Isin  (m — 4)(2n?r-(-x) 

a 

etc. 
Therefore,  if  we  put 

P?nn+x  =  cos  m  (2  n  K  -\-  x)  -\- m  cos  (m  —  2)  (2  n  ir  -f-  *)  +  etc. 
P'anTT+z  =  sin  m  (2  n  ?r  -f-  x)  -f  m  sin  (m  —  2)  (2  n  n  -f  x)  -f  etc. 
we  have 

(2  COS  X)m  =  Pto,  +  «  +  V  —  1  P'tHir+x  (<*) 

Now  m  being  a  fraction  (2  cos  x)M  has  imaginary  values,  but  when  cos  x  t's  positive,  it 
will  have  at  least  one  real  positive  value,  and  then  (2  cos  x)m  being  understood  to  de- 
note only  this  real  value,  all  the  values  are  included  in  the  formula 

(2  cos  x)m  X  (1)™  =  (2  cos  x)m  (cos  2  mn'tr  -)-  y  —  1  sin  2  mn'ir) 

*  Although  the  formulae  for  multiple  angles  require,  in  general,  the  combination  of 
two  series  when  m  is  not  an  integer,  yet  there  are  certain  cases,  even  when  m  is  a  frac- 
tion, in  which  one  or  tlie  other  of  the  series  will  disappear.  See  the  memoir  of  Poinsot, 
cited  at  the  beginning  of  this  chapter. 


142  PLANE  TRIGONOMETRY. 

Therefore  we  have 

(  2  cos  z)m  (cos  2  mn'  T  -f  i/  —  Isin2  mn'  n)  =  P2nir+I  -f  i/  —  1  -P'jnir+B 
Comparing  the  real  and  imaginary  terms, 

(2  cos  x)m  cos  2  mn'  K  =  P2nw+I 
(2  cos  x)m  sin  2  mn'  K  =  P'zmr+z 

and  to  find  the  corresponding  values  of  n  and  nx,  let  z  =  0,  then  (2  cos  x)m  =2m,  and 
the  series  become 

PJmr  =  cos  2mn7r  (1  +  m  +  m(m~1)  -f  etc.) 

m 

=  cos  2  mn  if  (1  -f-  I)1* 
=  2"1  cos  2  mn  TT 
and  in  the  same  way 

P/znir  =  2  m  sin  2  mn  TT 

Therefore  our  formulae  become 

2m  cos  2  win'  TT  =  2™  cos  2  mn  ir 
2™  sin  2  mnx  TT  =  2m  sin  2  mn  JT 

and  as  in  former  cases,  it  is  shown  that  n  =  n',  so  that  we  have  finally 

(2  cos  z)»  =    •**«**+--  (474) 


=n-^  (475) 

sin  2  mn  tr 

From  this  it  appears  that  the  real  and  positive  value  of  (2  cos  x)m  may  be  expressed 
either  by  a  series  of  cosines  or  by  one  of  sines  of  the  multiple  angles,  and  by  comparing 
(474)  and  (475),  we  have  the  following  constant  relation  between  these  series. 

P'tnir+x  _  sin  2  mn  TT 

Pjnir+z  COS  2  mn  7T 

245.  If  n  =  0,  (474)  gives 

(2  cos  x)m  =  Px  =  cos  mx  -f  m  cos  (m  -  2)  x  +  *»(*»—  ^  cos  (m  —  4)  x  +  etc.         (476) 

2 

which  may  be  employed  as  the  general  development  of  the  real  value  of  (2  cos  x)m, 
when  x  <  -• 

246.  The  same  supposition  of  n  =  0,  gives  sin  2  mn  n  =  0,  and  (475)  gives  there- 
fore, 

0  =  P'x  =  sin  mx  +  m  sin  (m  —  2)  x  +  "  (*»  —  1)  sjn  (m  _  4)  x  +  etc.  (477) 

2t 

a  remarkable  property  of  this  series  of  sines  of  multiple  arcs,  which  holds  for  all 

values  of  m,  provided  z  <  — 
2t 

247.  To  develop  any  power  of  the  cosine  of  the  simple  angle  in  a  scries  of  sines  or  cosine^ 
of  the  multiple  angles,  the  cosine  <>/  the  simple  angle  being  negative, 


MULTIPLE  ANGLES.  143 

If  the  denominator  of  n  is  even,  there  is  no  real  value  of  (2  cos  x)m  when  cos  x 
is  negative ;  but  we  may  put 

(2  cos  x)"  =  (—  2  cos  x)«  (—  !)•» 

=  (—  2  cos  x)m  [cos  m  (2  n'  +  1)  K  +  ^/  —  1  sin  m  (2  n'  +  1)  IT] 

which,  substituted  in  equation  (a)  of  Art.  244,  gives 

(— 2cosx)m  cosm  (2  n' +  1)  T  =  Pj»»  +  . 
(—  2  cos  x)m  sin  m  (2  n'  +  1)  TT  =  P'Swir  +  , 

Making  x  =  TT,  cos  x  =  —  1,  (—  2  cos  x)m  =  2™,  and  the  series  become,  by  the 
process  shown  in  Art.  244, 

•P(»»  +  ijir  =  2m  cos  TO  (2  n  -|-  1)  TT 

P/(ii«+)ir=2»  sinm  (2n  +  l)*r 
and  we  have 

2m  cos  TO(2n/-f-l)7r  =  2m  cos  m  (2  n  -f-  1)  w 
2m  sinm  (2  n' +  1)  IT  =  2™  sin  m  (2  n  -f  1)  ir 

whence,  as  before,  n  =  n',  and  our  formulae  are 

(—  2  cos  z)"  = J°!lnir  +  "      -  (478) 

cos  TO  (2«  +  l)  TT 

(-  2  cos  x)"  =  -     P^"^'     -  (479) 

sin  m  (2  n  +  1)  IT 

by  which  it  appears  that  the  real  value  of  ( —  2  cos  x)m  is  also  expressed  either  by 
a  series  of  cosines  or  of  sines  of  multiple  arcs,  which  series  have  the  constant  re- 
lation 

P'g  n  IT  +  x sin  m  (2  n  -\- 1)  TT 

Plnir  +  x        COS  m  (2  « -|-  1)  JT 

24ft.  If  n  =  0,  (478)  and  (479)  give 

(—  2  cos  x)«  =     P.*      = ~  (cos  rox-fmcos(ro  —  2)z  +  etc.)          (480) 

cos  m  TT       cos  m  TT 

(—  2  cos  x)m  =  ~x~  —  — - —  (sin  mx  +  m  sin  (m  —  2)  x  +  etc.)  (481) 

sin  m  TT       sin  TO  TT 

In  this  case  sin  m  TT  is  not  zero,  unless  m  is  an  integer,  so  that  the  series  Pfx  does 
not  become  zero  when  x  >  ~,  and  both  (480)  and  (481)  may  be  employed  as  the  true 

m 

developments  of  ( —  2  cos  x)m . 

249.  When  m  is  an  integer,  the  series  (476)  and  (480)  always  terminate  at  the 
(m  -f-  l)th  term  ;  and,  since  in  (480)  cos  nur  =  db  1,  according  as  TO  is  even  or  odd, 
and  ( —  2  cos  x)m  =  ±  (2  cos  z)m  in  the  same  cases,  both  (476)  and  (480)  become 

(2  cos  x)«  =  cos  mx  +  m  cos  (TO  —  2)  x  +  5^?— ill  cos  (m  —  4)  x  +  etc.     (482) 

2 

But  the  series  (481)  becomes  zero,  so  that  (482)  is  the  only  series  by  which 
(2  cos  x)m  can  be  developed  in  functions  of  the  multiple  arcs,  when  m  is  integral. 


144  PLANE  TEIGONOMETEY. 

250.  To  develop  any  power  of  the  sine  of  the  simple  angle,  in  a  series  of  sines  or  cosines 
of  the  multiple  angles. 

If  y  =  cos  x  -j-  y  —  1  sin  x,  we  have,  by  (435)  and  the  Binomial  Theorem, 
(l/  —  l)m  (2  sin  z)m  =  (y  —  y—  »)» 


_  ete> 

in  which  \f*,  ym~*,  etc.  have  the  same  values  as  in  Art.  244,  but  the  signs  of  the 
coefficients  are  alternately  +  *nd  —  »  so  that  if  we  put 

Qt*w  +  x  =  cos  m  (2  n  it  -f  x)  —  m  cos  (m  —  2)  (2  n  TT  -(-  x)  -f  etc. 

Q'inw  +  x  =  sin  m  (2  n  it  +  x)  —  m  sin  (m  —  2)  (2  n  TC  -f  x)  +  etc. 

we  have 

(V  —  l)m  (2  Bin  z)"  =$,„,  +  ,+  y  —  1  §'2nir  +  * 

Substituting  the  value  of  (^/  —  l)m  by  (446),  and  comparing  the  real  and  imaginary 
terms,  we  find 


(2  sin  ,)-  sin  '  =  V  ,.„  +  . 

8 

and  if  we  make  a:  =  -,  we  shall  find  by  the  process  frequently  employed  above 

m 

that  n  =  nx  ;  whence 

(2sinJe)»  =  -    -Slpi^  (483) 

cos  £  m  (4  n  -f-  1)  *r 

(2  sin  x)m  =  -  Q'*«*  +  *  —  (484) 

sin  £  m  (4  »  -f  1  )  ?r 

so  that  the  real  value  of  (2  sin  x)m  may  be  developed  in  either  the  cosines  or  sines  of 
the  multiples.    The  two  series  have  the  constant  relation 

g^ng^fs  _  sin  ft  m  (4  n  -f  1)  it 
Qtnn  +  x      cos  Jm  (4n  +  l)»r 

251.  If  n  =  0  in  (483)  and  (484), 
(2  sin  x)m  =  -  ^*  -  =  -  -  -  (cos  mx  —  m  cos  (m  —  2)  *  +  etc.)          (485) 


(2  sin  z)m  =  —  »-•  —  =  -r—  r  -  (sin  tnz  —  m  sin  (m  —  2)  x  +  etc.)          (486) 


both  of  which  series  are  applicable  when  m  is  fractional. 

252.  When  m  is  an  integer,  one  or  the  other  of  the  series  (485),  (486),  will  always 
be  zero,  according  to  the  form  of  m,  and  there  will  be  but  one  series  to  express 
(2  sin  x)m. 

If        m  —  4  m',  (2  sin  z)m  =        cos  mx  —  m  cos  (m  —  2)  a;  -f  etc.  (487) 

m  =  4  m'  -f  1,  (2  sin  z)w  =        sin  ma;  —  m  sin  (m  —  2)  z  +  etc.  (488) 

m  =  4  mr  +  2,  (2  sin  z)m  =  —  (cos  mx  —  m  cos  (m  —  2)  x  -f  etc.)  (489) 

m  =  4  m'  +3,  (2-sin  x)m  =  —  (sin  mz  —  m  sin  (m  —  2)  z  -f  etc.  )  (490) 


MULTIPLE  ANGLES.  145 

253.  The  series  (485)  and  (486)  become  zero  when  m  is  an  integer,  as  follows  : 
If  m  —  2m',  0  =  sinnu;  —  msin(m  —  2)  x  -f  etc.  (491) 


m  =z  2  m'  -f-  1,         0  =  cos  mx  —  m  cos  (m  —  2)  x  +  etc.  (492) 

The  reason  why  these  series  are  zero  is  obvious,  since  they  terminate  at  the 
(m  +  l)th  term,  the  terms  equally  distant  from  the  first  and  last  are  equal  with  oppo- 
site signs,  and  the  middle  term  of  (491)  is  zero. 

254.  Given  the  equation 

(493) 


to  express  x  ±  y  in  a  series  of  multiples  of  y. 
Substituting  the  values  of  tan  x  and  tan  y  given  by  (431) 


_ 

'      P 


-1  -f- 

whence 


or  putting 

fs=r£Zll  (494) 


e»  (z-v)V'-l  = 


Taking  the  Naperian  logarithms  of  both  members, 

2  (x  —  y)  i/  —  1  =  log  (1  —  q  e~»  J'*'-1)  —  log  (1  — 
and  developing  the  second  member  by  the  formula 

log  (1  —  n)  =  —  n  —  J  n*  —  J  ns  —  etc. 
we  have 

2  (2  _  y)  y  —  1  =  —  q  e-J  vv-1  _  £  9«  e-^^-i  —  J  g»  e-'vv'-1  —  etc. 


Substituting  in  the  second  member  by  (430), 

x  —  y  —  q  sin  2  y  -f  \  (f  sin  4  y  -f-  J  g*  sin  6  y  -f-  etc.  (6) 

The  equation  (a)  might  have  been  put  under  the  form 

1 L  £vV— i 


1  —  1- 

9 

from  which,  by  taking  the  logarithms  and  substituting  as  before, 

sin  2  y      sin  4  v      sin  6  y  /\ 

x  +  y  = *  — z z  —  etc.  (c) 

9  27s  Sg3 

In  this  investigation,  we  have,  in  effect,  used  Moivre's  formula,  in  its  limited  or 
less  general  form  ;  but  the  requisite  generality  may  be  given  to  our  results,  by  observ- 
ing, that  (493)  would   hold  if  we  were  to  substitute  tan  x  —  tan  (n'n  -f  *),  tan  y 
=  tan  (n/x7r  -(-  y),  and  therefore  we  may  substitute    for  the  first  member  of    (6), 
19  N 


146  PLANE  TRIGONOMETRY. 

n'Tr  -)-  z  —  (n"T  -(-  y)  =  x  —  y  —  (n"  —  n'}  it  =  x  —  y  —  wr,  n  being  (like  n'  and  n") 
an  arbitrary  integer  or  zero.  Hence,  the  required  general  development  of  z  —  y  in 
series  is 

z  —  y  =  nir  -f  q  sin  2  y  +  J  9*  sin  4  y  +  £  9*  sin  6  y  -f-  etc.  (495) 

In  like  manner,  since  tan  z  =  tan  (z  —  n/7r),  tan  y  =  tan  (y  —  TI^JT),  we  may  sub- 
stitute in  the  first  member  of  (c),  z  —  n'n  -\-  y  —  nf/^  =  x  -f-  y  —  w,  and  the  general 
development  of  z  -j-  y  in  series  is 

sin2y      sin  4  y      sin  6  y   ,  (AQC\ 

x  -f-  y  =  TIT  — * z- - £  -|-  etc.  (496) 

q  2q*  3  q3 

In  these  formulae  z  and  y  are  supposed  to  be  expressed  in  arc,  and  to  obtain  z  ^  y 
in  seconds,  the  terms  of  the  series  must  be  divided  by  sin  lx/. 

255.  The  preceding  problem  is  particularly  useful  in  finding  z  when  p  and  y  are 
given,  and  z  is  nearly  equal  to  y ;  in  which  case  p  is  nearly  equal  to  unity,  either 

q  or  —  is  a  small  fraction,  and  one  of  the  series  (495),  (496)  converges  rapidly. 

EXAMPLES. 

1.  Given  y  =  50°  and  p  =  1-00065,  to  find  z  from  (493). 
Taking  only  the  first  term  of  the  series  (495),  and  assuming  n  =  0, 

2  y  =  100°  log  sin  2  y    9'99335 

•00065  ,..  -„  7, 

a  = log  q     o  O1174 

2-00065 

ar  co  log  sin  1"    5-31443 

x  —  y  =  65"-995  log  (z  —  y)     1-81952 

z  =  50°  l'5"-995 

2.  Given  y  =  50°  and  p  =  — 1-00065,  to  find  z  from  (493).     In  this  case 

_  2-00065 
•00065 

and  the  computation  by  (496),  if  we  assume  n  =  0,  is 

2  y  =  100°  log  sin  2  y       9'99335 

_1=          '00065  loj-^l-  6-51174 


q  2-00065 

ar  co  log  sin  1"      5-31443 

z  -f  y  =  —  65"-995  log  (z  +  y)  —  1-81952 

z  =  —  y  —  65/A995  =  —  50°  I'  5 "'995 

or,  if  n  =  1,  z  =  180°  —  50°  1 '  5/A995  =  129°  58'  54/A005. 

In  general,  (493)  is  to  be  solved  by  (495)  when  p  is  positive,  and  by  (496)  when  p 
is  negative. 

256.  Given  the  equation 

sin  (a  -f-  2)  =wisinz  (497) 

to  express  z  in  a  series  of  multiples  of  a. 
We  deduce  as  in  Art.  168, 

an  (z  +  J  a)  _  ^— ^ 


MULTIPLE  ANGLES.  147 

which  is  reduced  to  (493)  by  putting 

x  =  z  +  $a  y  =  *«  p=  m_1 

whence  9  = 

p  -f- 1        m 

and  (495)  becomes 

i    sin  a    i    sin  Z>  a    ,   sin  o  ct     \  /  A(\Q\ 

z  =  HTT  H H -) +  etc.  (4yo) 

03  o  Vn3 

which  is  to  be  employed  when  m  >  1 ;  and  (496)  becomes 

z  -f-  a  =  TIT  —  m  sin  a  —  J  m*  sin  2  a  —  £  m*  sin  3  a  —  etc.  (499) 

which  is  to  be  employed  when  m  <  1,  n  being  any  integer  or  zero. 
257.  Given  the  equation 

(500) 


1  -(-  m  cos  a 


to  express  z  tn  a  series  of  multiples  of  a. 
This  equation  in  the  form 


sin  z  m  sin  a 


cos  z       1  +  m  cos  a 

gives  sin  z  -\-  m  sin  z  cos  a  =  m  cos  z  sin  o 

sin  z  =  m  sin  (a  —  z) 

tan  (a  —  J  a)  =  OT~1    tan  }  a 
m  -f  1 

which  is  reduced  to  (493)  by  substituting 


p  _ 
whence  g  =  i— 

p  +  1  m 

and  the  series  (495)  and  (496)  become 

(501) 


m  2  m?  3  wis 

z  —  n  TT  -(-  m  sin  a  —  J  m*  sin  2  a  +  J  m8  sin  3  o  —  etc.  (502) 

258.  Given  the  equation 

tanz=      msina  (503) 

1  —  m  cos  o 

to  express  z  in  a  series  of  multiples  of  a. 

The  equation  (500)  becomes  (503)  by  changing  the  signs  of  both  m  and  a:  the  same 
changes  in  (501)  and  (502)  give 

_etc.  (504) 


m  2  nr  3  m' 

=  n?r-|-Tnsin  a  -\-\rn*  sin  2  o  -|~  J  ms  sin  3  o  -f  etc.  (505) 


148  PLANE  TRIGONOMETRY. 

259.  In  a  plane  triangle  ABC,  given  a,  b  and  C,  to  find  A  or  B  by  a  series  of  multiples 
of  C. 
By  (260) 

~-  sin  C 

tan  A  = — 

1  —  -£  cos  (7 
b 

which,  compared  with  (503),  gives,  by  (505), 

(506) 


n  being  necessarily  =  0  in  this  case.     B  is  found  by  the  same  series,  interchanging 
a  and  b. 

260.  In  a  plane  triangle,  ABC,  given  a,  b  and  C,  to  find  c  by  a  series  of  multiples  of  C. 
We  have 

(507) 


4=  *       2*008(7+1 
a*        a1         a 

by  (451)  -(cosC  +  i/  —  lsinC)~| 

X  F      -  (cos  C—  i/  -  1  sin  C)~\ 
-£-=    [l-y  (cosC+i/  —  IsinC)]  X  [l  -  |-  (cos  (7-  T/  -  1  sin  <7)"| 

Taking  the  common  logarithms,  employing  in  the  second  member  the  formula 
log  (1  —  n)  =  —  M  (n  +  $  n»  +  J  n*  +  etc.) 

and  applying  Moivre's  Formula  (440)  in  expressing  the  powers  of  cos  0±  \/  —  1  sin  C, 
we  have 

2  log  c  —  2  log  6  = 

-M  Py  (cosC+V  —  lsin(7)  +  -^  (cos  2C-f  ^-1  sin  2C)  +etc.l 

—  JJf  P-  (cosC—  V  —  lsinC)+-^;r  (cos2C—  !/  —  1  sin  2  C)  -f  etc.] 
L  b  2o*  J 

i        i         vr  i  a  n   i     O1       COS  2  C    i     O*       COS  3  (7    ,       .     \  /CAO\ 

Iogc  =  log6  —  M  {—  coeC+  —  --  -  --  J-  —  •  —  ^  --  (-etc.j        (508) 

This  series  was  first  given  by  Legendre.  The  series  (495)  and  (496),  upon  which 
are  based  those  of  the  subsequent  articles,  (Arts.  256,  257,  258  and  259),  are  due  to 
Lagrange. 


PART  II. 


CHAPTER    I. 

GENERAL  FORMULAE. 

1.  SPHERICAL  TRIGONOMETRY  treats  of  the  methods  of  computing 
the  unknown  from  the  known  parts  of  a  spherical  triangle. 

It  is  shown  in  geometry,*  that  a  spherical  triangle  may,  in  gene- 
ral, be  constructed  when  any  three  of  its  six  parts  are  given,  (not 
excepting  the  case  where  the  three  angles  are  given).  We  are  now 
to  investigate  the  methods  by  which,  in  the  same  cases,  the  unknown 
parts  may  be  computed. 

We  shall  at  first  confine  our  attention  to  such  triangles  only  as  are 
treated  of  in  geometry,  namely,  those  whose  sides  are  each  less  than 
a  semicircumference,  and  whose  angles  are  each  less  than  two  right 
angles  ;  that  is,  those  in  which  every  part  is  less  than  180°. 

2.  It  is  shown  in  geometry,  that  if  a  solid  angle  is  formed  at  the 
center  of  a  sphere  by  three  planes,  the  three  arcs  in  which  these 
planes  intersect  the  surface  of  the  sphere  form  a  spherical  triangle. 
Now  the  real  objects  of  investigation  in  spherical  trigonometry  are 
the  mutual  relations  of  the  angles  of  inclination  of  the  faces  and 
edges  of  a  solid  angle ;  but,  for  convenience,  the  spherical  triangle 
which  forms  the  base  of  the  solid  angle  is  substituted  for  it.     The 
sides  of  the  triangle  being  proportional  to  the  angles  of  inclination 
of  the  edges  of  the  solid  angle,  are  taken  to  represent  those  angles ; 
and  the  angles  which  those  sides  form  with  each  other  are  regarded 


*  The  student  is  here  supposed  to  be  acquainted  with  Spherical  Geometry,  at  least 
so  much  of  it  as  is  to  be  found  in  Legendre's  treatise,  or  in  that  of  Prof.  Peirce,  of 
Harvard  University. 

N  2  149 


J50  SPHERICAL  TRIGONOMETRY. 

as  identical  with  the  angles  of  inclination  of  the  faces  of  the  solid 
angle.  But,  since  varying  the  radius  of  the  sphere  would  not,  in 
any  respect,  change  the  solid  angle,  or  the  values  of  the  angles  which 
enter  into  it,  the  mutual  relations  in  question  ought  to  be  deduced 
without  any  reference  to  the  magnitude  of  the  radius  of  the  sphere. 
In  fact,  we  shall  deduce  our  fundamental  formulae  from  a  direct  con- 
sideration of  the  solid  angle  itself. 

3.  In  a  spherical  triangle,  the  sines  of  the  sides  are  proportional  to 
the  sines  of  the  opposite  angles. 

FlG.  lt  Let  A  B  C,  Fig.  1,  be  a  spherical 

triangle,  0  the  center  of  the  sphere. 
The  angles  of  the  triangle  are  the 
inclinations  of  the  planes  A  0  B, 
A  0  C  and  B  0  C,  to  each  other,  and 
will  be  designated  by  A,  B  and  C; 
their  opposite  sides  respectively  will 
be  designated  by  a,  b  and  e,  as  in 
plane  triangles.  The  trigonometric 
functions  of  these  sides  will  be  the  same  as  those  of  the  angles 
B  0  C,  A  0  C,  A  0  B,  which  they  subtend  at  the  center  of  the  sphere. 
(PI.  Trig.  Art.  20.) 

From  any  point  B'  in  OB,  let  fall  B' P  perpendicular  to  the  plane 
AOC;  and  through  B' P  let  the  planes  B'PA',  Bf  P  C'  be  drawn 
perpendicular  to  0  A  and  0  C,  intersecting  the  plane  OAC  in  the 
lines  PA',  P  C',  and  the  planes  A  OB,  BOC  in  the  lines  A'  B',  B'C'. 
The  plane  triangles  A'PBf,  B' PC'  are  right  angled  at  P •  and 
OA'B',  0  C'B'  are  right  angled  at  A'  and  C'.  The  angle  B'  A'  P, 
being  formed  by  two  lines  perpendicular  to  OA,  is  the  measure  of 
the  inclination  of  the  planes  AOB,  AOC,  or  of  the  angle  A',  and 
B'  C'  P  is  the  measure  of  the  angle  C. 
We  have  therefore,  by  PI.  Trig.  Art.  15, 


B'  P 
sin  A  =  sin  B'A'P=      ~ 


whence 


sin  A        B'P        Bf  C'   _  B[C'  } 

tinC      B'A'       IV  P       IV  A' 


GENERAL  FORMULAE. 


151 


Again, 


sin  c  =  sin  B'OA'  = 


B'A' 
B'O 


whence 


sin  a 
sin  c 


B'O       B'C' 


B'O       B'A'       B'A' 


Comparing  (m)  and  (n), 


sin 


sin  A 

sin  c       sin  C 


(n) 


(1) 


which  in  the  form  of  a  proportion  is 

sin  a  :  sin  c  —  sin  A  :  sin  C 

which  is  the  theorem  that  was  to  be  proved. 

4.  In  Fig.  1,  A,  a,  Cand  c,  are  each  less  than  90°,  but  the  con- 
struction would  not  vary  if  any  of  these  parts  were  greater  than  90°, 
except  that  the  points  A'  and  C'  might  be  found  in  the  lines  AO,  CO, 
produced  through  0  ;  and  one  or  more  of  the  right  triangles  A'B'P, 
etc.,  would  contain  the  supplements  of  A,  a,  C,  or  c  instead  of  these 
quantities  themselves.     But  the  sine  of  an  angle  and  of  its  supple- 
ment being  the  same,  the  preceding  demonstration  would  still  be 
valid,  so  that  the  theorem  is  applicable  to  any  spherical  triangle. 

Indeed,  according  to  PI.  Trig.  Art.  49,  this  result  follows  from 
the  nature  of  the  trigonometric  functions  themselves,  and  the  demon- 
stration of  the  preceding  theorem  might  therefore  be  considered  as 
general,  without  requiring  a  special  examination  of  the  various  posi- 
tions of  the  lines  of  the  diagram. 

5.  In  a  spherical  triangle,   the  cosine  of  any  side  is  equal  to  the 
product  of  the  cosines  of  the  other  two  sides,  plus  the  continued  product 
of  the  sines  of  those  sides  and  the  cosine  of  the  included  angle. 

Let  the  plane  B'A'C',  Fig.  2,  be 
drawn  perp.  to  O  A,  intersecting  the 
planes  AOB,  BOC  and  AOC1,  in  the 
lines  A'B',  B'  C'  and  A'C'.  Then  the 
angle  B'A'C'  =  A,  and  B'OC'  =  a, 
and  by  PI.  Trig.  Art.  119,  in  the 
triangles  A'  B'C',  OB'C',  we  have 


FIG.  2. 


B'C"*  =-•  A'B'2  +  A'C'2  —  2  A'B'  . 
B'C'2  =  OB'2  +  O  C'2  -  2  O  B'  . 


A'C'  cos  A 
0  C'  cos  a 


152  SPHERICAL  TRIGONOMETRY. 

Subtracting  the  first  of  these  equations  from  the  second,  and  observ- 
ing that  in  the  right  triangles  OA'B',  OA'C', 

O  B'2  —  A'  Bn  =  0  A'2,        0  C"2  —  A'  C'2  =  O  A'2 
we  have 

0  =  2  OA'2  +  2  A'B' .  A'C'  cos  A  —  2  0  Br .  0  C'  cos  a 

OA' .  OA'       A'B'  .A'C' 
whence         cos  a  =  Q£f    Q  ^  +  Q£,    ^  cos  A 

Substituting  the  trigonometric  functions  derived  from  the  right  tri- 
angles OA'B',  OA'C', 

cos  a  =  cos  6  cos  c  -J-  sin  6  sin  c  cos  A  (2) 

which  is  the  theorem  to  be  proved.  It  may  be  regarded  as  the  fun- 
damental theorem,  for  the  preceding  (1)  can  be  deduced  from  it,  but 
as  the  process  is  somewhat  circuitous,  we  have  preferred  deducing 
the  two  theorems  from  independent  constructions. 

6.  In  the  construction  of  Fig.  2,  both  6  and  c  are  supposed  less 
than  90°,  while  no  restriction  is  placed  upon  A  and  a  •  but  the  equa- 
tion (2)  is  no  less  applicable  to  all  the  other  cases  if  the  principle  of 
PI.  Trig.  Art.  49  be  granted.  As  that  principle  may  not  be  suffi- 
ciently evident  to  the  student  unacquainted  with  analyti- 
cal geometry,  we  shall  verify  it  in  this  case,  as  follows.* 
1st.  In  the  triangle  ABC,  (Fig.  3),  let  b  <  90°  and 
\c  c>  90°.  Produce  BA,  BCto  meet  in  B',  forming  the 
lune  BB' ;  then  AB'  —  180°  —  c,  and  6  are  both  <  90°, 
and  the  preceding  demonstration  would  apply  to  the 
triangle  AB'C.  Therefore,  applying  (2)  to  AB'C,  we 
have 

cos  (180°— a)=cos  6  cos  (1 80°  —  c)  +  sin  6  sin  (180°—  c)  cos  (180°— 4) 
or  by  PI.  Trig.  (64), 

—  cos  a  =  —  cos  b  cos  c  —  sin  b  sin  c  cos  A 
and  changing  all  the  signs 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A 

the  same  result  that  would  have  been  found  by  applying  (2)  directly 
to  ABC. 

*  Hymer's  Spherical  Trigonometry.     Cambridge,  1841. 


GENERAL  FORMULA. 


153 


2d.  In  the  triangle  ABC,  Fig.  4,  let  b  >  90°,  c  >  90° ; 
produce  AB  and  AC  to  meet  in  A' ;  then  A  'B  and  A'C 
being  both  less  than  90°,  the  formula  (2)  is  applicable  to 
A' EG.  Therefore 

cos  a  =  cos  (180°  —  6)  cos  (180°  —  c) 

+  sin  (180°  —  6)  sin  (180°  —  c)  cos  A 
=  ( —  cos  6)  ( —  cos  c)  -f  sin  b  sin  c  cos  A 
=  cos  6  cos  c  +  sin  6  sin  c  cos  A 


the  same  result  as  before. 

7.  The  theorems  expressed  by  (1)  and  (2)  being  applied  succes- 
sively to  the  several  parts  of  the  triangle,  give  the  two  following 
groups : 

sin  a  sin  B  =  sin  6  sin  A 

sin  b  sin  C  =  sin  c  sin  B  >      (3) 

sin  c  sin  A  =  sin  a  sin  C 

cos  a  =  cos  6  cos  c  +  sin  6  sin  c  cos  .4 

cos  6  =  cos  c  cos  a  -\-  sin  c  sin  a  cos  J?  f      (4) 

cos  c  =  cos  a  cos  6  +  sin  a  sin  6  cos  C 


8.  Let  A'B'C',  Fig.  5,  be  the  polar  triangle 
of  ABC,  and  designate  its  angles  and  sides  by 
A',  B',  C',  a',  b'  and  c'.  Then,  by  geometry, 


a' =  180°  — 4 
6'  =  180°  —  B 
c'  =  180°  —  C 


^'  =  180°  —  a, 
5'  =  180°  -  6, 
C'  =  180°  —  c, 

and  applying  the  first  equation  of  (4)  to  A'B'C', 
cos  af  =  cos  b'  cos  c'  -f-  sin  b'  sin  c'  cos  A 
or  by  PL  Trig.  (64), 

—  cos  A  =  (—  cos  B)  ( —  cos  (7)  +  sin  B  sin  C  (—  cos  a) 

—  cos  .4  =  cos  B  cos  C —  sin  B  sin  G  cos  a 

Changing  the  signs  of  this,  we  have  the  first  of  the  following  group : 


cos  A  =  —  cos  B  cos  (7+  sin  B  sin  Ccos  a 
cos  B  =  —  cos  C  cos  A  +  sin  C  sin  A  cos  b 
cos  0=  —  cos  A  cos  j5  +  sin  A  sin  5  cos  c 


(6) 


20 


154  SPHERICAL  TRIGONOMETRY. 

It  is  thus  that,  by  means  of  the  polar  triangle,  any  formula  of  a 
spherical  triangle  may  be  immediately  transformed  into  another,  in 
which  angles  take  the  place  of  sides,  and  sides  of  angles. 

9.  Several   other   important  fundamental  groups  of   formula?  are 
obtained  from  the  preceding  with  the  greatest  ease. 

The  first  of  (4)  multiplied  by  cos  c  is 

cos  a  cos  c  =  cos  6  cos2  c  -f-  sin  6  sin  c  cos  c  cos  A 
and  the  second  of  (4)  is  the  same  as 

cos  a  cos  c  +  sin  a  sin  c  cos  B  =  cos  b 
the  difference  of  which  is 

sin  a  sin  c  cos  B  =  (1  —  cos2  c)  cos  b  —  sin  6  sin  c  cos  c  cos  A 
Since  1  —  cos'c  =  sin2  c,  this  may  be  divided  by  sin  c,  and  gives 

sin  a  cos  B  =  sin  c  cos  b  —  cos  c  sin  6  cos  A 
whence         sin  b  cos  C=  sin  a  cos  c  —  cos  a  sin  c  cos  B  (6) 

sin  c  cos  A  =  sin  6  cos  a  —  cos  6  sin  a  cos  C 

If  we  interchange  jB  and  C,  and  therefore  also  b  and  c,  the  group 
becomes 

sin  a  cos  C  =  sin  b  cos  c  —  cos  b  sin  c  cos  A 

sin  6  cos  A  =  sin  c  cos  a  —  cos  c  sin  a  cos  J5  V      (7) 

sin  c  cos  />  =  sin  a  cos  6  —  cos  a  sin  6  cos  C 

10.  If  (6)  and  (7)  are  applied  to  the  polar  triangle,  they  give,  after 
changing  the  signs  of  all  the  terms, 

sin  A  cos  6  =  sin  C  cos  B  -\-  cos  C  sin  B  cos  a 
sin  7?  cos  c  =  sin  J.  cos  C  -f  cos  ^4  sin  C  cos  6  I      (g) 

sin  C  cos  a  —  sin  B  cos  .4  +  cos  B  sin  ^4  cos  c 
and 

sin  ^4  cos  c  =  s\n  B  cos  C  -f-  cos  B  sin  6'  cos  a 

sin  B  cos  a  =  sin  Ccos  A  -f-  cos  (7 sin  ^4  cos  6  j>      (9) 

sin  Ccos  b  =  sin  A  cos  5  -f~  cos  A  sin  J?  cos  c 

11.  Dividing  the  first  of  (6)  by  the  following  derived  from  (3), 

sin  a  sin  B        ,     , 
— r          -  =  sin  6 
sin  A 


GENERAL  FORMULAE.  155 

we  find  the  first  of  the  following  group 

sin  A  cot  B  —  sin  c  cot  6  —  cos  c  cos  A  ^j 

sin  jB  cot  C  =  sin  a  cot  c  —  cos  a  cos  B  >     (10) 

sin  C  cot  A  =  sin  6  cot  a  —  cos  6  cos  C  } 

and  in  the  same  way  from  (7),  or  by  interchanging  the  letters  B  and 
(?,  b  and  c  in  (10),  we  find 

sin  A  cot  C  =  sin  b  cot  c  —  cos  6  cos  ^1  ^ 

sin  B  cot  .A  =  sin  c  cot  a  —  cos  c  cos  .B  >     (11) 

sin  C  cot  jB  =  sin  a  cot  6  —  cos  a  cos  (7  J 

If  (10)  are  applied  to  the  polar  triangle,  we  find  (11),  so  that  no 
new  relations  are  elicited. 

12.  The  preceding  formulae  are  sufficient  to  furnish  a  theoretical 
solution  for  every  case  of  spherical  triangles,  but  some  transforma- 
tions are  required  to  facilitate  their  application  in  practice. 

In  the  first  of  (4)  substitute,  by  PL  Trig.  (1 39), 

cos  A  =  l  —  2  sin2  £  A 
we  find,  by  PI.  Trig.  (39), 

cos  a  =  cos  (6  —  c)  —  2  sin  b  sin  c  sin2  £  A  (12) 

and  we  have  similar  expressions  for  cos  b  and  cos  c. 
If  we  substitute  in  (4),  by  PL  Trig.  (138), 

cos  A  =  —  1  -f-  2  cos2  £  A 
we  find,  by  PL  Trig.  (38), 

cos  a  =  cos  (b  +  c)  -f-  2  sin  b  sin  c  cos2  J  A  (13) 

and,  of  course,  similar  expressions  for  cos  b  and  cos  c. 

13.  Substituting  in  (5) 

cos  a  =  1  —  2  sin2  J  a  =  —  I  +  2  cos2  ^  a 
we  find  by  the  same  process 

cos  A  =    -  cos  (B  +  C)  —  2  sin  B  sin  C  sin2 1  a          (14) 
cos  ^4  =   -  cos  (j5  —  C)  +  2  sin  5  sin  C  cos2  }  a          (15) 

which  might  have  been  obtained  by  applying  (12)  and  (13)  to  the 
polar  triangle. 


156  SPHERICAL  TRIGONOMETRY. 

14.  If  in  ( 12)  we  substitute  cos  o  =  1  —  2  sin*  J  a,  cos  (b  —  c)  =  1  —  2  sin*  J  (6  —  c), 
we  obtain  the  first  of  the  following  equations  ;  and  the  others  are  obtained  by  a  similar 
process  from  (12j,  (13),  (14)  and  (15). 

sin*  j  a  =  sin"  J  (b  —  c)  +  sin  b  sin  c  sin*  J  A  (16) 

sin*  |  o  =  sin*  £  (6  -f  c)  —  sin  6  sin  c  cos*  $  ^4  (17) 

cos*  £  a  =  cos2  J  (6  —  c)  —  sin  b  sin  c  sin*  J  A  (18) 

cos*  J  o  =  cos*  £  (6  +  c)  -f-  sin  6  sin  c  cos*  £  .4  (19) 

sin*  }  A  =  cos*  J  (B  +  C )  -f  sin  B  sin  C'  sin*  J  a  (20) 

sin*  }  A  =  cos*  i  (5  -  6')  —  sin  B  sin  C  cos2  $  a  (21 ) 

cos*  i  .-1  =  sin*  J  (J3  +  C)  —  sin  £  sin  C  sin*  £  o  (22) 

cos1  M  =  sin*  }  (JS— C)  +  sin  £sin  (7  cos*  *  a  (23) 

15.  By  PL  Trig,  we  have 

1  =  cos*  J  A  +  sin*  }  ^4 
cos  A  =  cos*  i  ^4  —  sin*  J  .4 
whence 

cos  b  cos  c  =  cos  6  cos  c  cos*  J  A  +  cos  6  cos  c  sin*  f  .4 

sin  6  sin  c  cos  ^4  =  sin  6  sin  c  cos*  \  A  —  sin  6  sin  c  sin*  \  A 
the  sum  of  which  is,  by  (4), 

cos  o  =  cos  (6  —  c)  cos1  \  A  -f-  cos  (6  +  c)  sin1  J  A  (24) 

and  substituting  1  —  2  sin*  \  a,  etc.,  for  cos  a,  etc. 

sin*  }  a  =  sin*  $(&  —  «)  cos*  }  ^  +  sin*  J  (6  +  c)  sin1  J  A         (25) 
cos*  i  a  =  cos*  }  (6  —  c)  cos*  ±A  +  cos*  J  (6  +  «)  sin*  i  ^        (26) 

la  the  same  manner  we  deduce  from  (5) 

cos^  =  — cos(B  — C)sin»  Jo  — cos  (£  +  <?)  cos*  }  a  (27) 

sin»M=       cos*H-B-C')sin'ia  +  cos*J(JS  +  (7)cos'ia  (28) 

cos1  M  =       8>n2  J  (-P  -  C1)  si"11  i  «  +  »'"'  *(&  +  <?)  cos*  J  a  (29) 

It  is  hardly  necessary  to  add  that  each  of  the  equations  (12  to  29)  gives  a  group  of 

three,  by  applying  it  successively  to  the  three  sides  or  three  angles  of  the  triangle. 


16.  From  (12)  we  find 
sin2  \ 
If,  in  PI.  Trig.  (108),  we  put 


cos  (6  —  c)  —  cos  a 

sin2  i  A  = •-— ; — r-r^ 

2  sin  b  sm  c 


x  =  a,  y  =  b—  c 

whence 

H*  +  y)  =  H°  +  6-c)»         H*-y)  =  i  («-&  +  <») 

we  find 

cos  (b  —  c)  —  cos  a  =  2  sin  £  (a  —  b  -f-  c)  sin  %  (a  -f  b  —  c) 
which,  substituted  in  the  above  equation,  gives 

.  ,  ,  sin  4  (a  —  6  -f  c)  sin  ^(a  -\-b-c) 

sin*  +  A  =  -  .      — ; — *-* — 

sin  o  sm  c 


GENEKAL  FORMULAE. 


157 


Let  s  denote  the  half  sum  of  the  sides,  that  is,  let 

a-j-6  +  c  =  2s,  £  (a  -{-  &  -f  c)  =  * 

then 

a—  6  +  c  =  a  +  6  +  c—  2  6  =  2  s  —  26  =  2  («  —  6) 
a-\-  b  —  c  =  a-\-b  +  c  —  2c  =  2s— 2c  =  2(«  —  c) 

which  substituted  in  (30)  give 

.  ,  ,  sin  (s  —  6)  sin  (s  —  c) 

sm2 1  A  =  - 

sin  6  sm  c 

.  ,  .   n      sin  (s  —  c)  sin  (s  —  a) 

whence  also          smz  \  B  =  —      — ; — *• — r-)l £ 

sm  c  sm  a 


sin 


»  in—  sin  (s  —  a)  sin  (s  —  6) 
$  o  —  —          ; . 

sm  a  sm  6 


17.  From  (13)  we  find 

cos2  \  A  = 


(31) 


2 .    ,  cos  a  —  cos  (6  -4- 


2  sin  6  sin  c 

and  from  PI.  Trig.  (108),  by  making 

x  —  6  +  c,  y  — 


we  find 

cos  a  —  cos  (6  +  c)  —  2  sin  |-  (a  +  6  +  c)  sin  ^  (6  +  c  —  a) 
which,  substituted  above,  gives 

+  ^  +  c)  sin  ^  (6  +  c  —  a) 


cos2  4  A  = 


sn 


sin  6  sin  c 
Introducing,  as  in  the  preceding  article,  s  =  ^  (a  +  6  -f-  c), 


cos 


2  ,    .  sin  s  sin  (s  —  a) 

sin  6  sin  c 


cos 


D       sin  s  sin  (s  —  6) 

Jo  = — 


sm  c  sm  a 


cos 


2  j  ~ sin  g  sin  (s  —  c) 

sin  a  sin  6 


(33) 


158  SPHERICAL  TRIGONOMETRY. 

18.  The  quotient  of  (31)  divided  by  (33)  gives 

tan2  %A  =  sin('s~6)sin(*-c) 
sin  s  sin  (s  —  a) 

f;qn2i    K  _  sin  (s  —  c)  sin  (s  —  a) 
tan  TJ-  j_>  —        .         .  —  -        —  — 
sin  s  sin  (s  —  6) 

,  ,  r,      sin  (s  —  a)  sin  (s  —  6) 
tan2  i(?=-. 

sin  s  sin  (s  —  c) 


19.  From  (14)  we  find 


Bin  *  a  =  — 


cos  A  +  cos  (B  -|-  <?) 


2  sin  B  sin  (7 
from  which,  by  PI.  Trig.  (107),  we  deduce 


sn  %  a 
and  if  we  put 


cos 


cos 


sin      a  .  = 


sin  j5  sin  C 


—  cos  S  cos  ($  — 
- 

sin  B  sin  C 


.  »  ,  ,        —  cos  S  cos  (S  —  J5) 
sin2  A  b  =  — 

sin  C  sin  A 


—  cos  <S  cos  (S  —  C) 
sin2  4-  c  =  - 

sin  ^4  sin  lj 


(34) 


(35) 


(36) 


The  first  member  of  each  of  these  equations  being  a  square,  the 
second  member  must  be  essentially  positive,  although  its  algebraic 
sign  is  negative;  in  fact,  since  by  geometry  2  £>180°,  $>90°, 
cos  S  is  negative,  and  —  cos  S  is  positive. 

20.  From  (15)  we  find 


,  ,  cos  A  -f  cos  (B  —  C} 

cos2 1  a  =  - 


2  sin  B  sin  C 
from  which  we  deduce,  by  a  process  similar  to  the  preceding, 


cos*  £  a  = 


r>    •     n 
sin  B  sin  C 


GENERAL  FORMULA. 


159 


COS 


ai      _cos(S—B)cos(S  —  C) 
i  -5-  a  —  . 

sin  E  sin  6 


cos 


cos 


2  j  , cos  (8  —  C)  cos  (ff  —  A) 

sin  C  sin  A 

_cos(S—A)cos(S—B) 

I       If    C    .  J          •  T-» 

sin  J.  sin  B 


(38). 


21.  From  (36)  and  (38) 


2  1  —  cos  8  cos  (/$  —  J.) 

tan  -a-  <z  —  ~ 

cos  (S  —  J5)  cos  (S  —  C) 

2  IT,-  —  COS  S  COS  (ff  —  -B) 

~  " 


2  l     = 


(39) 


We  might  have  deduced  (36),  (38),  (39),  by  applying  (31),  (33), 
(34)  to  the  polar  triangle. 

22.  Napier's  Analogies.     Dividing  the  1st  of  (34)  by  the  2d,  we 
find 

tan  ^  A sin  (s  —  6) 

tan  ^  B       sin  (s  —  a) 


Regarding  this  as  a  proportion,  we  have,  by  composition  and  divi- 
sion, 

tan  ^  A  -f-  tan  ^  B sin  (s  —  b)  -f-  sin  (s  —  a) 

tan  ^  A  —  tan  ^  B      sin  (s  —  6)  —  sin  (s  —  a) 

In  PI.  Trig.  (109),  if  we  put  x  =  s  —  6,  y  =  s  —  a,  whence 


(m) 


x  ~H  V  —  2s  —  a  —  b  =  c 
x  —  y  =  a  —  b 


we  have 


sin  (s  —  b)  +  sin  (s  —  a) 


tan 


sin  (s  —  6)  —  sin  (s  —  a)        tan  %  (a  —  6) 
and  by  PI.  Trig.  (126), 

tan  $  A  +  tan  ±  B  ^  sin  $  (A  +  B) 
tan  \  A  —  tan  %  B  ~  sin  ^  (A  —  B) 


160  SPHERICAL  TRIGONOMETRY. 

Therefore  (m)  becomes 

sin  ^(A  +  B)  _        tan^c 
sin^(A  —  B)      tan  %  (a  —  b) 

or       sin  £  (A  -f  -B)  :  sin  ^(A  —  -B)  —  tan  \  c  :  tan  \  (a  —  6) 

which  is  the  first  of  Napier's  Analogies. 

23.  Again,  the  product  of  the  1st  and  2d  of  (34)  gives 

,  „       sin  (s  —  c) 
tan  £  A  tan  ^  £  =  - 

sin* 

or  1  :  tan  £  ^4  tan  ^  .B  =  sin  s  :  sin  (s  —  c) 

whence,  by  composition  and  division, 

1  —  tan  \  A  tan  ^  B  _  sin  s  —  sin  (s  —  c)  *  . 

1  -f  tan  ^  A  tan  %  B      sin  s  -f  sin  (s  —  c) 

By  PI.  Trig.  (109),  if  x  =  *,  y  —  s  —  c,  we  have 

sin  s  —  sin  (a  —  c)  _        tan^e 
sin  s  -f  sin  (s  —  c)      tan  ^  (a  +  6) 

and  by  PI.  Trig.  (127), 

1  —  tan  4-  A  tan  ^  B  =  cos  |  (A  +  B) 
I  +  tan  $  A  tan  £  B  ~  cos  \(A  —  B) 

Therefore  («)  becomes 


cos^(A-B)      tan  $  (a  +6)  (41) 

or    cos  $(A  +  B)  :cos£(^  —  B)  =  tan  £  c  :  tan  £  (a  -f  6) 


which  is  the  second  of  Napier's  Analogies. 

24.  If  (40)  and  (41)  are  applied  to  the  polar  triangle,  we  shall  find 

sin  ^  (a  4-  b)  _        cot^  C 

sin  £  (a  —  6)  ~  tan  %(A  —  B)  I    (42) 

or     sin  \  (a  -j-  6)  :  sin  ^  (a  —  6)  =  cot  ^  C :  tan  \  (A  —  B) 
cos  ^  (a  +_6)  _        cot  ^  C 

;  (43) 

or     cos  -^  (a  +  6)  :  cos  ^  (a  —  6)  =  cot  ^  (7 :  tan  ^  ( A  +  B) 
which  are  the  third  and  fourth  of  Napier's  Analogies. 


GENERAL  FORMULAE.  161 

25.  Gauss's  Theorem.    If 

p  =  cos  i  c  sin  £  (A  -f  JB)  P=  cos  J  (7  cos  J  (a  —  6) 

5  =  cos  £  c  cos  J  (>i  +  •#)  Q  —  si°  £  C'cos  i  (a  +  6) 

r  =  sin  £  c  sin  i  (.4  —  JB)  J2  =  cos  £  Csin  J  (a  —  6) 

«  =  sin  J  c  cos  i  (  .4  —  .Z?)  5  =  sin  $  C  sin  £  (a  +  6) 

<Aen  <Ae  products  p  X  9,  P  X  r,  p  X  »>  9  X  r,   ?  X  «,  *"  X  *,  are  respectively  equal  to  the 

products  PXQ,PXR,PXS,QXR,QXS,RXS. 

First.  From  (3)  we  have 

sin  c  (sin  ^4.  :fc  sin  J5)  =  sin  C  (sin  a  ±  sin  6) 
which,  by  PI.  Trig.  (105),  (106)  and  (135),  are  reduced  to 

sin  Jccos  Jcsin  J  (A  +  -B)  cos  J  (  A  —  J5)  =siu  J  C'cos  \  Csin  \  (a  -\-b)  cos  J  (a  —  6) 
sin  Jccosjccos  J(^4  +  J5)  sin  £  (4  —  JS)  =sinj  Ccos  J  Ccos  J  (a  +  6)  sin  J  (a  —  6) 

or 

ps  =  PS  and    qr=QR 

Second.  From  (6)  and  (7) 

sin  c  (cos  JB  ±  cos  A)  =  (1  =F  cos  C)  sin  (a  ±  b) 
which,  by  PI.  Trig,  are  reduced  to 

sin  Jccosjccos  J  (A  +  J5)  cos  J  (A  —  _B)=sin  J  (7sin  J  (7sin  J  (a  -j-  6)  cos  J  (a  +  6) 
sin  J  c  cos  J  c  sin  %  (A  -\-  B)  sin  J  (^4  —  JB)  =  cos  ^  Ccos  J  (7  sin  i  (a  —  6)  cos  J  (a  —  b) 

or 

qs—QS  and    pr  —  PR 

Third.  From  (8)  and  (9) 

(1  ±  cos  c)  sin  (A  ±  J5)  =  sin  C  (cos  6  ±  cos  a) 
which,  by  PI.  Trig,  are  reduced  to 


cos  J  c  cos  J  c  sin  J  (  .4  +  J5)  cos  J  (-4  +  J5)  =  gin  i  C'cos  J  C'cos  J  (a  +  &)  cos  J  (a  —  6) 
sin  }  c  sin  J  c  sin  J  (^4  —  J5)  cos  $  (.4  —  J5)  =  sin  J  (7  cos  i  (7  sin  J  (a  -f-  6)  sin  J  (a  —  6) 
or 

pq  =  PQ  and     rs  =  RS 

26.   TAe  notation  of  the  preceding  article  being  still  employed,  the  quantities  p*t  q*t  rf,  «*, 
are  respectively  equal  to  Ps,  ^2,  fi*,  S*. 
We  have 


and  qr  —  Q  R 

the  quotient  of  which  is 

p*  =  P1      whence  p  =  db  P 
and  in  the  same  way  q*  =  Q2  g  =  ±  Q 


21  o2 


162  SPHERICAL  TRIGONOMETRY. 

27.  In  these  last  equations,  the  positive  sign  must  be  used  in  all  the  second  members,  or  the 
negative  sign  in  all  of  them.     For  if  we  take 

P  =  +  P 
the  equations 

pq  =  PQ,        pr  =  PE,        ps  =  PS 
being  divided  by  this,  give 

q=+Q,          r=  +  R,         s  =  +  S 

and  if  we  take 

p  =  -P 

the  same  equations,  divided  by  this,  give 

q  =  —Q,        r  =  —  R,        s  =  —  S 
We  have  therefore  the  following,  which  are  generally  cited  as  Gauss's  Equations. 


cos  \  c  sin  J  (A  -)-  B)  =  cos  \  C  cos  £  (a  —  6) 
cos  \  c  cos  £  ( A  +  B)  =  sin  \  C  cos  \  (a  -f-  b) 
sin  £  c  sin  £  (4  —  E)  =  cos  £  (7  sin  £  (a  —  6) 
sin  \  e  cos  ^  (A  —  B)  =  sin  ^  C  sin  £  (a  -f-  b) 

cos  |  c  sin  J  (.4  +  .B)  =  —  cos  J  (7  cos  J  (a  —  6) 
cos  £  c  cos  £  (A  -f-  5)  =  —  sin  J  C  cos  J  (a  -f-  6) 
sin  J  c  sin  J  (J.  —  B)  =  —  cos  $  C  sin  £  (a  —  b) 
sin  J  c  cos  \  (-4  —  -B)  =  —  sin  £  (7  sin  J  (a  +  6) 


(44) 


(45) 


If,  however,  we  consider  only  those  triangles  whose  parts  are  all  less  than  180°,  the 
first  of  these  groups,  (44),  is  alone  applicable,  for  we  must  then  have  p  =  -\-  P;  since 
cos  £  c,  sin  £  (A  +  £),  cos  £  C,  cos  J  (a  —  6)  are  then  all  positive  quantities.  The  use 
of  (45)  will  be  seen  in  the  chapter  on  the  solution  of  the  general  spherical  triangle. 

Napier's  Analogies,  (40),  (41),  (42)  and  (43)  can  be  deduced  directly  from  (44). 

ADDITIONAL  FORMULAS. 

28.  We  shall  here  add  some  formulae  which,  though  not  so  frequently  used  as  the 
preceding,  are  either  remarkable  for  their  elegance  and  symmetry,  or  of  importance  in 
certain  inquiries  of  astronomy  and  geodesy. 

29.  The  product  of  (30)  and  (32)  gives 

.  .  A       4  sin  s  sin  (s  —  a)  sin  («  —  6)  sin  (a  —  c)  (AR^. 

sm  A  — '       .  (to) 

sin1  o  sin'  e 

Put 

n*  =  sin  s  sin  (s  —  a)  sin  (s  —  6)  sin  (s  —  c)  (47) 

then  sin  A  =  - — — - 

sin  o  sin  c 

and  in  the  same  manner 


(48) 

2n 
sin  B  =  — 

sin  a  sin  c 

the  quotient  of  which  is 


sin  A  _sin  a 
sin  B      sin  6 

which  is  our  first  theorem,  Art.  3.  As  (48)  was  obtained  from  (30)  and  (32),  and  these 
from  (4)  without  the  aid  of  (3),  we  may  consider  the  whole  fabric  of  spherical  trigo- 
nometry as  resting  upon  the  fundamental  formulae  (4). 


ADDITIONAL  FORMULAE. 
30.  We  have  also  from  (35)  and  (37) 

gin2  o  ^  —  4  cos  S  cos  (S—  A)  cos  (S -  B)  cos  (8—  C) 
and  if 


sin2  B  sin2  C 


*  =  —  cosS cos  (S—  A)  cos  (S—  B)coa(S  —  O) 

2N 


From  (48)  and  (51), 


n 

N 


sin  B  sin  O 

sin  a sin  6  sin  c 

sin  A       sin  B      sin  (7 


163 

(49) 

(50) 
(51) 

(52) 


31.  If  we  develop  (47)  and  (50)  by  PI.  Trig.  (173)  and  (174) 

4  ri*   =1  —  cos2  a    —  cos*  6   —  cos2  c   -{-  2  cos  a  cos  6  cos  c 
4  jY2  =  1  —  cos"  A  —  cos2  B  —  cos2  C  —  2  cos  A  cos  B  cos  C 


(53) 
(54) 


32.  The  following  simple  results  are  easily  deduced  from  the  equations  (31  to  38) 

cos  \  A  cos  ^  B sin  a      . 

sin  J  0  sin  c 


$  ^  sin  $  .B  _  sin  (s  —  a) 
cos  J  C  sin  c 


sn 


cos 


(55) 


(56) 


cos  £  (7 

sin  |  .4  sin  ^  J?  _  sin  (s  —  c) 
sin  \  C  sin  c 

sin  \  a  sin  |  6  _  —  cos  S 
cos  £  c  sin  (7 

sin  ^  a  cos  |  6  _  cos  (S  —  A) 
sin  \  c  sin  C 

cos  |  a  sin  \  b  _  cos  (S  —  B) 
sin  ^  c  sin  C 

cos  |  a  cos  ^  6  _  cos  (S  —  C) 
cos  ^  c  sin  C 

33.  By  means  of  (55)  and  (56)  we  can  deduce  expressions  for  the  functions  of 
s,  s  —  a,  etc.,  in  terms  of  the  angles,  or  of  S,  S  —  A,  etc.,  in  terms  of  the  sides.  We 
have,  from  (51), 

sinc  =  _  2JV        =  _  N  _ 
sin  A  sin  B      2  sin  £  A  cos  \  A  sin  \  B  cos  J  B 

which,  substituted  in  (55),  gives 

sin  s  = 


N 


sin  (s  —  c)  = 


2  sin  £  A  sin  ^  B  sin  £  (7 

N 


2  cos  i  .4  cos  £  £  sin  J  C 
whence,  by  interchanging  the  letters,  we  have  also  sin  (s  —  a)  and  sin  («  —  b). 


(57) 
(58) 


164  SPHERICAL  TRIGONOMETRY. 

Again,  we  have 

sin  (a  —  e)  =  sin  s  cos  c  —  cos  s  sin  c 
whence 


_  sin  s  cos  c  —  sin  (s  —  c) 

,  — * 1— 


cos  s- 

sin  c 


which,  by  (55),  is  reduced  to 

cos  i  A  cos  i  B  cos  c  —  sin  i  A  sin  4  B 

cos  s  = * * « * — 

sin  $  (7 

and  from  the  equation 

cos  (s  —  c)  =  cos  s  cos  c  +  sin  s  sin  c 
we  find,  by  substituting  (55)  and  (59), 

™  f,      »\  _  —sin  fr  ^4  sin  J  B  cos  c  +  cos  J  -4  cos  i  S 
sinTo — 

To  eliminate  c  from  the  second  members  of  (59)  and  (60),  we  have,  by  (58) 

cos  C  +  cos  A  cos  .B 

sin  A  sin  5 
whence 

cos  J  .4  cos  J  B  cos  c  =  • 


:    i   /<  „•     ID  cos  (7  +  cos  A  cos 

sin  J  .4  sin  }  /j  cos  c  =  --  !  —  - 

$  A  cos  %  B 


2  sin  i  a  sin  $  6  cos  $  c 
.     o  _  sin  \  a  sin  ^  6  cos  C  +  cos  ^  a  cos  }  fe 


which,  substituted  in  (59)  and  (60),  give 

_  cos  A  -\-  cos  B  -f-  cos  C  —  1  _  1  —  sin'  \  A  —  sina  \  B  —  sin*  ^  C  //,.  ^ 

4  sin  J  A  sin  J  £  sin  JO  2  sin  £  .4  sin  £  £  sin  J  (7 

,  __  v  _  cos^t-j-cos/?  —  oos(74-l_  c°sa  |  A  -|-  cos'  ^  P  —  cos'  J  (7  /go) 

= 


From  the  preceding  we  easily  deduce 

tan  ,  =  _  2JV  _  =  _  siiLC  _  (63) 

cos  A  +  cos  B  +  cos  (7  —  1      cos  c  —  tan  J  A  tan  £  B 

tan  (s  —  c)  ==  _  2  N  __  =  _  8m  c  _  .  (64) 

COt  %A  COS  J  5  —  COS  C 


34.  The  equations  (57  to  64)  applied  to  the  polar  triangle,  give, 

_cosS=-  -  —"  3-  (65) 

2  cos  ^  o  cos  ^  6  cos  i  c 

(66) 


ADDITIONAL    FORMULAE.  165 

.        ]  +  cos  a  -f-^osj>  j=_cog_c  _  cos2  £  a  -f  cos2  £  6  +  cos2  £  c  ^j_          ^  ^ 

4  cos  J  a  cos  £  6  cos  \  c,  2  cos  \  a  cos  £  6  cos  £  c 

/  o      ,r<\       cos  i  a  cos  J  b  cos  C+  sin  j  «  sin  fr  6  /fiq\ 

sin  ( o  —  C/ 1  ==  V"*/ 

cos  j  c 

•  <-v_  (71  — 1  ~ cos  a  ~~  c°gA+-g°jLg  —  gin!-Lg-+ sin*  3 fc  ~ sinit  ^ c      (70) 

4  sin  i  a  sin  £  6  cos  £  c    "        2  sin  i  a  sin  |  6  cos  J  c 

2» sin  C  (71) 

~  I  -j-  cos  a  +  cos  b  -j-  cos  c   cos  (7  -f-  cot  J  a  cot  J  6 

.,«•.>«  2n  sin  O  ^ 


'       1  —  cos  o  —  cos  6  +  cos  c      cos  C  +  tan  $  a  tan  J  6 
35.  From  (68)  we  find 

.     o      2  cos  i  a  cos  i  6  cos  i  c  —  cos"  \  a  —  cos"  £  b  —  cos"  £  c  +  1       ,7o\ 

1 — sino  =  —  ,  , ; —  (i*) 

2  cos  5  a  cos  £  o  cos  £  c 

-    .     .     „ 2  cos  }  a  cos  ^  6  cos  \  c  +  cos"  |  a  +  cos2  ^  6  +  cos8  ^  e  —  1       ,_^ 

2  cos  £  a  cos  J  6  cos  \  c 

the  numerators  of  which  may  be  reduced  by  PI.  Trig.  (173)  and  (174),  by  making 
x  =  |a,  j/  =  J6,  2  =  Jc,  whence  v  =  }  (a  -f  6  +  c)  =  \  s,  v  —  x  =  \  (s  —  o),  etc. : 
therefore, 

..         .     ^  _  2  sin  £  s  sin  %  (s  —  a)  sin  %  (s  —  b)  sin  |  (s  —  c)  ,7-^ 

J.  —  Sill  o  —  -  -tit  \       ) 

cos  £  a  cos  £  o  cos  }  c 

..    ,     .      o 2  cos  |  s  cos  fr  (s  —  a)  cos  £  (a  —  b)  cos  £  («  —  c)  /-g-, 

cos  J  a  cos  \  b  cos  J  c 

The  product  of  these  equations  reproduces  (65) ;  their  quotient  is,  by  PI.  Trig.  (154), 

tan2  (45°  —  J  S)  =  tan  J  s  tan  J  (s  —  a)  tan  J  (s  —  6)  tan  J  (s  —  c)  (77) 

36.   CagnoWs  Equation. — Multiplying  the  first  equation  of  (4)   by  cos  A,  we  find 

cos  a  cos  A  =  cos  b  cos  c  cos  A  -f  sin  6  sin  c  —  sin  b  sin  c  sin*  A 
and  from  (5)  in  a  similar  manner, 

cos  a  cos  A  =  —  cos  B  cos  C  cos  a  -f-  sin  B  sin  0  —  sin  B  sin  (7  sin1  o 

Observing  that  by  (3)  we  have  sin  6  sin  c  sin2  A  =  sin  B  sin  C  sin2  a, 
these  two  equations  give, 

sin  6  sin  c  -f  cos  b  cos  c  cos  A  =  sin  5  sin  (7 —  cos  B  cos  (7 cos  a  (78) 

a  relation  between  the  six  parts  of  the  triangle,  first  given  by  CAGNOLI.  It  is  a 
property  of  this  equation  that  either  member  is  a  function  which  Acw  the  same  value  in  a 
given  spherical  tnangle  and  its  polar  triangle.  Thus,  if  we  distinguish  the  sides  and 
angles  of  the  polar  triangle  by  accents,  we  have* 

sin  6  sin  c  -{-  cos  b  cos  c  cos  A  =  sin  &'  sin  c'  -(-  cos  bf  cos  c'  cos  Af  (79) 

*  See  Mathematical  Monthly,  (Cambridge,  Mass.,)  vol.  i.  p.  282. 


166  SPHERICAL    TRIGONOMETRY. 

37.   To  deduce  the  formulae  of  plnne  triangles  from  those  of  spherical  triangles. 

The  analogy  of  many  of  the  preceding  formulae  with  those  of  plane  triangles  is 
sufficiently  obvious.  We  can,  in  fact,  deduce  the  plane  formula;  from  those  of  this 
chapter,  by  regarding  the  plane  triangle  as  described  upon  a  sphere  whose  radius  is  in- 
finite, the  triangle  being  an  infinitely  small  portion  of  the  sphere.  The  quantities  a,  b  and 
c  must,  in  this  case,  express  the  absolute  lengths  of  the  sides  ;  and  the  angles  which 

they  subtend  at  the  center  of  the  sphere,  expressed  in   arc,  will  be     ,    — ,    — ,  r  be- 

r       r      r 

ing  the  radius  of  the  sphere.     When  r  is  very  large,    — ,    — ,   — ,     are    very    small, 

r       r      r 

and  we  may  express  the  values  of  sin  — ,  cos — ,   etc.    approximately,   by  one   or  two 

r          r 

terms  of  their  expansions  in  series,  PI.  Trig.  (405)  and  (400),  and  if  their  values  be 
substituted  in  our  spherical  formulae,  we  shall  obtain  approximate  relations  between  the 
sides  and  angles  of  the  triangle.     If  we  then  make  r  infinite  we  shall  obtain  exact 
relations  between  the  sides  and  angles  of  a  plane  triangle. 
Thus  we  have 

.  a    a    a8  a5 

sin  — h  etc.   a  — h  etc. 

sin  A      r    r   2.3  rs  2.3  r2 


sin  B   in_6.   1 ^  +  etc.   b  -  -*-  +  etc. 

r    r   2.3  r3  2.3  r3 

and  making  r  infinite,  we  find  the  formula  of  PI.  Trig. 

sin  A  ji 

sin  B        b 

In  the  same  manner 


cos  A  = 


a  b         c        -,        a2     ,  / ,         62         c2     ,   62  c2         ,    \ 

cos cos  —  cos  —  — \-  etc.  —  (1  —  —  etc.  ] 

r  r          r  _  2  r2  I  2?-*       2r*        4r*          _f 


.     b      .     c 

sin  —   sin  — 

r  r 


-     +  etc. 


and  making  r  infinite,  we  have  the  formula  of  PI.  Trig. 


c  —  a 

~r~~ 

2  be 


Formulae  that  involve  only  the  sines  or  tangents  of  the  sides  may  be  reduced  im- 
mediately to  the  plane  formula?  by  substituting  a,  6,  etc.,  for  sin  a,  tan  a,  etc.  Thus, 
(31  to  34)  give  the  corresponding  formula?  of  PI.  Trig,  by  omitting  the  symbol  sin.  ; 
and  (40),  (41),  by  omitting  the  symbol  tan.  when  these  symbols  are  prefixed  to  sides. 


CHAPTER    II. 

SOLUTION  OF  SPHERICAL  EIGHT  TRIANGLES. 

38.  WHEN  one  of  the  angles  of  a  spherical  triangle  is  a  right 
angle,  the  general  formulae  of  the  preceding  chapter  assume  forms 
that  are   remarkably  analogous  to  the  relations  established   for  the 
solution  of  plane  right  triangles,  and  equally  simple  in  their  appli- 
cation. 

39.  Let  C=9Q°,  Fig.  6.     From  (3)  we  Fl0-6- 
have 

.      ,       sin  a    .     ~ 

sm  A  =  —. —  sm  C 
sin  c 

but  since  (7=  90°,  sin  C—  1 ;  therefore, 


sin  A  = 


and,  in  the  same  manner, 


sin  J5  = 


sin  a 
sin  G 


sin  b 


sm  c 


(80) 


that  is,  the  sine  of  either  oblique  angle  of  a  spherical  right  triangle  is 
equal  to  the  quotient  of  the  sine  of  the  opposite  side  divided  by  the  sine 
of  the  hypotenuse.     Compare  PI.  Trig.  (1). 
40.  From  (11),  we  find 


cos  A  = 


sin  b  cot  c  —  sin  A  cot  C 
cos  6 


but  if  0=  90°,  cot  C=  0  ;  therefore, 


sin  b  cot  c 
cos  A  =  -  -  =  tan  b  cot  c 


cos  b 


or 


cos  A  = 


tan  b 
tan  c 


COS  ±f  = 


tan  a 
tan  c 


(81) 


167 


168  SPHERICAL  TRIGONOMETRY. 

that  is,  the  cosine  of  either  angle  is  equal  to  the  tangent  of  the  adjacent 
side,  divided  by  the  tangent  of  the  hypotenuse.  Compare  PI.  Trig.  (1). 

41.  From  (10),  we  have, 

sin  b  cot  a  —  cos  b  cos  O 

cotA  =  -  .    n • 

sin  C 

which,  when  C—  90°,  becomes 

.    ,  sin  b 

cot  A  =  sin  b  cot  a  =  — 

tana 

or,  taking  the  reciprocals, 

A       tan  a  ^       tan  b  /oox 

tan  A  =  -  tan  B  =  -  (82) 

sin  6  sin  a 

that  is,  the  tangent  of  either  angle  is  equal  to  the  tangent  of  the  opposite 
side,  divided  by  the  sine  of  the  adjacent  side.  Compare  PI.  Trig.  (1). 

42.  From  (5),  we  find, 

cos  B  -\-  cos  C  cos  A 

sm  A  = . • 

cos  6  sm  C 

and  if  C=  90°, 

cos  B  .     „      cos  A  /eQ>. 

sm  A  =  —  sm  B  =  —  (83) 

cos  b  cos  a 

that  is,  the  cosine  of  either  angle,  divided  by  the  cosine  of  its  opposite 
side,  is  equal  to  the  sine  of  the  other  angle.  In  PL  Trig,  we  have 
sin  A  =  cos  B. 

43.  From  (4),  we  have, 

cos  c  =  cos  a  cos  b  -f-  sin  a  sin  b  cos  C 
or,  when  C=  90°, 

cos  c  =  cos  a  cos  b  (84) 

that  is,  the  cosine  of  the  hypotenuse  is  equal  to  the  product  of  the  cosines 
of  the  two  sides.  In  PL  Trig,  c2  =  a2  +  b2. 

44.  From  (5), 

cos  C  +  cos  A  cos  B 
cos  c  =  -    — —  - — — TT- 
sm  yl  sm  75 

or,  when  0=90°, 

COS  J.  COS  J?  ,,,  /QKx 

cos  c  =  —  -—  —  cot  -4  cot  B  (85) 

sm  J.  sm  B 


SOLUTION  OF  SPHERICAL  EIGHT  TRIANGLES. 


169 


that  is,  the  cosine  of  the  hypotenuse  is  equal  to  the  product  of  the  cotan- 
gents of  the  two  angles.     In  PI.  Trig.,  1  =  cot  A  cot  B. 

45.  No  difficulty  will  be  found  in  remembering  the  preceding  for- 
mulae for  spherical  right  triangles,  if  they  are  associated  with  the 
corresponding  ones  for  plane  triangles :  thus, 


In  plane  right  triangles. 

In  spherical  right  triangles. 
.      .       sin  a       •     r>      8in  ^ 

sin  A  sin  B  —  = 
c                           c 

sin  A.  —   .           sin  Jj  —   . 
sm  c                     sin  c 

A       tan  6            D      tana 

cos  A  —  -            cos  B  — 
c                          c 

!„,,      4           a                      4011    P           " 

CuS  A  —                     COS  JL>  — 

tan  c                    tan  c 
A       tana            D      tan& 

tan  A  —                tan  Jo  — 
b                           a 

tan  A  —   .    .      tan  Jo  —   . 
Bin  6                     sin  a 

.     A      cos  B     .     jj      cos  A 

sin  A  -  —  cos  B,     sin  B  —  cos  A 

c*  =  az  +  b2 
1  =  cot  A  cot  B 

sin  A  -  —  •              sin  JJ  — 
cos  6                    cos  a 

cos  c  =  cos  a  cos  6 
cos  c  =  cot  A  cot  B 

46.  Napier's  Rules.  By  putting  these  ten  equations  under  a  different  form,  Napier 
contrived  to  express  them  all  in  two  rules,  which,  though  artificial,  are  very  generally 
employed  as  aids  to  the  memory. 

In  these  rules,  the  complements  of  the  hypotenuse  and  of  the  two  oblique  angles 
are  employed  instead  of  the  hypotenuse  and  the  angles  themselves.  The  right  angle 
not  entering  into  the  formula,  they  express  the  relations  of  five  parts,  but  in  the  rules 
the  five  parts  considered  are  a,  b,  co.  c,  co.  A  and  co.  B.  Any  one  of  these  parts  being 
called  a  middle  part,  the  two  immediately  adjacent  may  be  called  adjacent  parts,  and 
the  remaining  two,  opposite  parts.  The  right  angle  not  being  considered,  the  two  sides 
including  it  are  regarded  as  adjacent  parts.  The  rules  are : 

I.  The  sine  of  the  middle  part  is  equal  to  the  product  of  the  tangents  of  the  adjacent 
parts. 

II.  The  sine  of  the  middle  part  is  equal  to  the  product  of  the  cosines  of  the  opposite  parts. 
The   correctness  of  these  rules  will  be  shown  by  taking  each  of  the  five  parts  as 

middle  part,  and   comparing  the  equations   thus  found  with  those   already  demon- 
strated. 

1st.  Let  co.  c  be  the  middle  part ;  then  co.  A  and  co.  B  are  the  adjacent  parts,  o  and 
b  the  opposite  parts,  and  the  rules  give 


sin  (co.  c)  =  tan  (co.  A)  tan  (co.  B) 
sin  (co.  c)  =  cos  o  cos  6 


or      cos  c  =  cot  A  cot  B 
cos  c  =  cos  a  cos  & 


which  are  (85)  and  (84). 

2d.  Let  co.  A  be  the  middle  part;  then  co.  c  and  b  are  the  adjacent  parts,  co.  B  and 
o  the  opposite  parts,  and  the  rules  give 


sin  (co.  A)  =  tan  (co.  c)  tan  6 
sin  (co.  A)  =  cos  (co.  B)  cos  o 


cos  A  =  cot  c  tan  b 
cos  A  =  sin  B  cos  a 


22 


170  SPHERICAL  TRIGONOMETRY. 

In  the  same  manner,  if  co.  B  is  taken  as  the  middle  part, 

sin  (co.  B)  =  tan  (co.  c)  tan  a  or      cos  B  =  cot  c  tan  a 

sin  (co.  B)  =  cos  (co.  A)  cos  6  cos  B  =  sin  A  cos  b 

and  these  four  equations  are  the  same  as  (81)  and  (83). 

3d.  Let  o  be  the  middle  part ;  then  co.  B  and  6  are  the  adjacent  parts,  co.  A  and 
co.  c  the  opposite  parts,  and  the  rules  give, 

sin  a  =  tan  (co.  B)  tan  b  or      sin  a  =  cot  B  tan  6 

sin  a  =  cos  (co.  A)  cos  (co.  c)  sin  a  =  sin  A  sin  c 

In  the  same  manner,  if  b  is  taken  as  the  middle  part, 

sin  b  =  tan  (co.  A)  tan  a  or      sin  b  =  cot  A  tan  o 

sin  b  =  cos  (co.  B)  cos  (co.  c)  sin  b  =  sin  B  sin  c 

and  these  four  equations  are  the  same  as  (80)  and  (82). 

It  appears,  therefore,  that  these  rules  include  all  the  ten  equations  previously 
proved ;  and  they  include  no  others,  since  we  have  taken  each  part  successively  as 
the  middle  part. 

In  the  application  of  these  rules,  it  is  unnecessary  to  use  the  notation  co.  A,  co.  B, 
co.  c,  since  we  may  write  down  at  once  sin  A  for  cos  (co.  A ),  etc.* 

47.  In  order  to  solve  a  spherical  right  triangle,  two  parts  must 
be  given,  and  from  the  equations  of  Art.  45,  that  equation  must  be 
selected  which  expresses  the  relation  between  these  two  parts  and  the 
required  part. 

When  Napier's  Rules  are  employed,  it  is  only  necessary  to  determine  which  of  the 
three  parts — the  two  given  and  the  one  required — is  to  be  taken  as  the  middle  part. 
"These  three  parts  are  either  all  adjacent  to  each  other,  in  which  case  the  middle  one 
is  taken  as  the  middle  part,  and  the  other  two  are  adjacent  parts;  or  one  is  separated 
from  the  other  two,  and  then  the  part  which  stands  by  itself  is  the  middle  part,  and 
the  other  two  are  opposite  parts."  f 

48.  In  order  to  distinguish  the  functions  of  parts  less  than  90°  from 
those  greater  than  90°,  it  will  be  necessary  carefully  to  observe  their 
algebraic  signs,  according  to  PI.  Trig.  Art.  40.    But  when  a  required 
part  is  determined  by  its  sine,  since  the  sine  of  an  angle  and  of  its 
supplement  are  the  same,  there  will  be  two  angles,  both  of  which 
may  be  regarded  as  solutions,  except  when  this  ambiguity  is  removed 
by  either  of  the  following  principles. 

*  If  we  employ  as  the  five  parts,  the  hypotenuse,  the  two  angles,  and  the  comple- 
ments of  the  two  sides  including  the  right  angle,  these  parts  will  be  the  complements 
of  those  used  in  Napier's  Rules,  and  we  shall  have 

MAUBUIT'S  RULES. — I.  The  cosine  of  the  middle  part  is  equal  to  the  product  of  the 
cotangents  of  the  adjacent  parts. 

II.  The,  cosine  of  the  middle  part  is  equal  to  the  product  of  the  sines  of  the  opposite  parts. 

With  a  little  attention  at  the  commencement,  however,  and  by  observing  the  ana- 
logy exhibited  in  Art.  45,  the  student  will  find  that  he  will  have  little  use  for  either 
of  these  artificial  rules. 

t  Peirce's  Spherical  Trigonometry. 


SOLUTION  OF  SPHERICAL  RIGHT  TRIANGLES.  171 

49.  In  a  right  spherical  triangle,  an  angle  and  its  opposite  side  are 
always  in  the  same  quadrant,  that  is,  either  both  less  or  both  greater 
than  90°.  For,  by  (83), 

cos  B 


sin  A  = 


cos  b 


in  which,  since  sin  A  is  always  positive,  (A  <  180°),  cos  B  and  cos  b 
must  have  the  same  sign ;  that  is,  £  and  6  must  be  either  both  less 
or  both  greater  than  90°. 

50.  When  the  two  sides  including  the  right  angle  are  in  the  same 
quadrant,  the  hypotenuse  is  less  than  90°,  and  when  the  two  sides  are  in 
different  quadrants,  the  hypotenuse  is  greater  than  90°.     For,  by  (84) , 

cos  c  =  cos  a  cos  6 

in  which,  if  a  and  b  are  in  the  same  quadrant,  cos  a  and  cos  6  have 
like  signs,  and  cos  c  is  positive,  that  is,  c  <  90° ;  but  if  a  and  6  are 
in  different  quadrants,  cos  a  and  cos  6  have  different  signs,  and  cos  c 
is  negative,  that  is,  c  >  90°. 

We  proceed  now  to  the  solution  of  the  several  cases. 

51.  CASE  I.     Given  the  hypotenuse  and   one  angh,  or  c  and  A, 
Fig.  6. 

To  find  a.  The  relation  among  the  three 
parts,  c,  A,  and  a,  (as  in  PI.  Trig,  with  the 
same  data),  is  given  by  the  sine  of  A ;  and 
by  Art.  45, 


FIG.  6. 


B 


,       sm  a 

sm  A  =  — 

sm  c 

from  which  we  find* 

sin  a  =  sin  c  sin  A  (86) 

There  will  be  two  values  of  a  corresponding  to  the  same  sine,  but, 
by  Art.  49,  the  true  value  is  that  which  is  in  the  same  quadrant 
as  A. 

To  find  b.  The  relation  among  the  three  parts,  c,  A,  and  b,  (as  in 
PI.  Trig,  with  the  same  data),  is  given  by  the  cosine  of  A,  or, 

tan  b 
cos  A  =  - 

tan  c 

from  which  t  tan  b  —  tan  c  cos  A  (87) 

*This  equation  would  be  found  by  Napier's  Rules,  taking  a  as  the  middle  part, 
t  We  find  the  same  result  by  Napier's  Rules,  taking  co.  A  as  the  middle  part. 


172  SPHERICAL  TRIGONOMETRY. 

To  find  B.     We  have,  by  (85),* 

cos  c  =  cot  A  cot  B 

COS  C 

from  which  cot  B  =  —    -  =  cos  c  tan  A  (88") 

cot^l 

The  quadrants  in  which  b  and  B  are  to  be  taken,  will  be  deter- 
mined by  means  of  the  signs  of  tan  6  and  cot  B}  according  to  PI. 
Trig.  Art,  40. 

Check.  To  guard  against  numerical  errors,  it  is  often  expedient 
to  compute  the  same  quantity  by  two  different  and  independent 
methods.  In  many  cases,  however,  we  may  test  the  accuracy  of 
several  operations  by  a  single  formula,  which  may  be  called  the  check. 
In  the  present  instance,  when  the  three  parts,  a,  6,  and  J3,  have  been 
found,  we  should  have,  by  (82),  the  relation 

sin  a  =  tan  b  cot  B 
so  that  if  the  work  is  correct,  we  shall  find 

log  sin  a  =  log  tan  6  +  log  cot  B 

EXAMPLES. 

1.  Given  c  =  110°  46'  20",  A  =  80°  10'  30",  to  solve  the  triangle. 

By  (86).  By  (87).  By  (88). 

c,  log  sin  9-9708106     log  tan  —  0-4210061          log  cos  —  9*5498045 
Ay  log  sin  9-9935833     log  cos  +  9-2320794          log  tan  +  07615038 

log  sin  a  9-9643939  log  tan  b  —  9-6530855      log  cot  B—  0-3113083 

log  tan  6  —  9-6530855 

Check,     log  sin  a  +  9-9643938 

-4ns.  a  =  67°  6'  52"-7,     b  =  155°  46'  42"-7,     B=  153°  58'  24"-5 

2.  Given  c  =  120°,  A  =  120°  ;  solve  the  triangle. 

Ana.  a  =  131°24'34"-7       6  =  40°  53'  36"-2       .5  =  49°  6' 23"-8 

52.  If  A  =  90°,  we  must  also  have,  by  (85),  c  =  90°,  and  then 
tan  6  =  ^  tan  S  =  ~- 

eo  that  b  and  B  are  both  indeterminate;  that  is,  there  is  an  indefinite  number  of  tri- 
angles which  satisfy  the  given  values  of  c  and  A  ;  but  since 

cos.B  =  cos6sin.4=cos& 
we  always  have  B  =  b',  and  since 

sin  a  =  sin  c  sin  A  =  1 
we  have  a  =  90°,  and  all  the  parts  of  the  triangle  are  equal  to  90°,  except  b  and  B. 

If  only  c  is  given  =  90°,  all  the  parts  of  the  triangle  are  equal  to  90°,  except  A 
and  a ;  and  we  have  A  —  a. 

*  Or  by  Napier's  Rules,  taking  co.  c  as  the  middle  part. 


SOLUTION  OF  SPHERICAL  EIGHT  TRIANGLES.  173 

53.  CASE  II.   Given  the  hypotenuse  and  a  side,  or  c  and  a. 
To  find  A.     We  have  by  (80), 

sin  a  .  /onx 

sm  A  =  — —  =  cosec  c  sin  a  (89) 

sin  c 

To  find  B.     By  (81), 

cos  B  =  —  -  =  cot  c  tan  a  (90) 

tanc 

To  find  b.     By  (84), 

cos  c  =  cos  a  cos  6 


from  which  cos  b==—    -'=  cos  c  sec  a  (91) 

cos  a 

Check.     We  have  between  A,  B,  and  6,  the  relation 
cos  B  =  sin  A  cos  6 

EXAMPLES. 

1.  Given  c  =  140°,  a  =  20° ;  solve  the  triangle. 

By  (89).  By  (90).  By  (91). 

c,  log  cosec  0-1919325      log  cot  —  0-0761865      log  cos  —  9-8842540 
a,     log  sin  9-5340517      log  tan  +  9*5610659      log  sec  +  0-0270142 
log  sin  A  9-7259842  log  cos  B  —  9-6372524   log  cos  b  —  9-9112682 

log  sin  A  -f  9-7259842 

Check,     log  cos  B  —  9-6372524 
Ans.     A  =   32°    8'  48"-l 
J?  =  115°  42'  23"-8 
b  =  144°  36'  28"-4 

2.  Given  c  =  101°  16'  16"'7,     b=  115°  42'  38"-5;  find  A. 

Ans.     A  =  65°  32'  56"'4 

54.  When  a  =  c  and  consequently  both  =  90°,  sin  A  =  1,  .4  —  90°,  and 

0  0  T> 

~"0  ~~0 

so  that  B  =  b,  but  both  are  indeterminate  as  in  Art.  52. 

p  2 


174 


SPHERICAL  TEIGONOMETRY. 


55.  CASE  III.   Given  one  angle  and  its  opposite  side,  or  A  and  a. 
We  shall  have 


sin  A  = 


tan  A  = 


sin  ±1  == 


sm  a 
sin  c 

tan  a 
sin  6 

cos  A 
cos  a 


whence    sin  c  =  cosec  A  sin  a  (92) 

./ 
sin  6  =  cot  J[  tan  a  (93) 

sin  J5  =  cos  A  sec  a  (94) 

Check,     sin  6  =  sin  c  sin  jB 


In  this  case,  there  are  always  two  solutions,  all  the  required  parts 
being  determined  by  their  sines,  and  the  ambiguity  not  being  removed 
by  either  Art.  49  or  Art.  50.  This  also  appears  from  Fig.  7. 

If  AB  and  AC  be  produced  to  meet  in  A',  ABA'  and 
AC  A'  are  semicircumferences  and  A  =  A'  ;  the  triangles 
ABC  and  A'BC  both  contain  the  given  parts  A  and  a, 
c  but  c',  b'  and  B'  are  respectively  the  supplements  of  c, 
b  and  B.  It  must  not  be  inferred  that  in  every  case  all 
the  required  parts  are  less  than  90°  in  one  triangle,  and 
greater  than  90°  in  the  other;  but  the  proper  values  for 
each  triangle  must  be  selected  by  Arts.  49  and  50. 

EXAMPLES. 


1.  Given  A  =  100°,  a  —  112°  ;  solve  the  triangle. 


Am.     c=    70°  18'  10"-2 

6  =  154°    7'26"-5 

£  =  152°  23'    l"-3 


or 


c=109°41' 

b=    25°  52' 
B=    27°  36' 


2.  Given  A  =  80°,  a  =  68°;  solve  the  triangle. 

Ans.     c=    70°  18'  10"-2     ^  f     c  =  109°  41' 

b=    25°52'33"-5      V     or      I 


=     27°  36'  58"-7 


6  =  154°    7' 
B  =152°  23' 


3.  Given  B  =  150°,  6  =  160°;  solve  the  triangle. 

Ans.     c  =  136°  50'  23"'3     >,  f     c=     43°    9' 

a=    39°    4'  50"-7      V  or      I      a  =140°  55' 

^1=    67°    9'42"-7     J  i    4  =  112°  50' 


49"-8 

33"-5 

58"-7 


49"*8 

26"-5 

1"'3 


9"'3 
17"-3 


SOLUTION  OF  SPHERICAL  RIGHT  TRIANGLES.  175 

56.  CASE  IV.    Given  one  angle  and  its  adjacent  side,  or  A  and  b. 
We  shall  find  the  required  parts  by  the  equations 

cos  B  —  sin  A  cos  b  (95) 

tan  a  =  tan  A  sin  b  (96) 

cot  c  =  cos  A  cot  b  (97) 
Check,     cos  B  =  tan  a  cot  c 

EXAMPLES. 

1.  Given  ^.  =  80°  10'  30",  6  =  155°  46'  42"-7;  solve  the  triangle. 

Ans.  5  =  153°  58' 24"-5 
a=  67°  6'52"-6 
c=110°  46'  20"-0 

2.  Given  5  =  152°  23'  l"-3,  a=  112°  0'  0";  solve  the  triangle. 

Ana.     A  =  100° 

b  =  154°    7'  26"-5 
c=    70°  18'  10"-2 

57.  CASE  V.   Given  the  two  sides,  a  and  b. 
We  find  the  required  parts  by  the  equations 

cos  c  =  cos  a  cos  b  (98) 

cot  A  =  cot  a  sin  b  (99) 

cot  5  =  sin  a  cot  6  (100) 
Check,     cos  c  =  cot  .4  cot  5 

EXAMPLE. 
Given  a  =  116°,  6  =  16°;  solve  the  triangle. 

4n*.  fe  =  114°  55'  20"-4 
A=  97°  39'  24"-4 
£  =  ••  17°  41'39"-9 

58.  CASE  VI.     Given  the  two  angles,  A  and  B. 
The  required  parts  are  found  by  the  formulae 

cos  c  =  cot  A  cot  J5  (101) 

cos  a  =  cos  A  cosec  J?  (1 02) 

cos  b  =  cosec  A  cos  5  (103) 
Check,     cos  c  =  cos  a  cos  b 


176  SPHERICAL  TRIGONOMETRY. 

EXAMPLE. 

Given  A  =  60°  47'  24"'3,  B  =  57°  16'  20"-2;  solve  the  triangle. 

Ans.  c  =  68°  56'  28"'9 
a  =  54°  32'  32"-l 
b=  51°43'36"-1 

ADDITIONAL  FORMULA  FOB  THE  SOLUTION  OF  SPHERICAL  RIGHT  TRIANGLES. 

59.  As  in  plane  trigonometry,  cases  occur  in  which  particular  solutions  of  greater 
accuracy  than  the  ordinary  ones  are  required.     (PI.  Trig.  Art.  112.) 

60.  From  (89)  we  find 

1  —  Bin  A sin  c  —  sin  a 

1  +  sin  A      sin  c  -\-  sin  a 

which  by  PL  Trig.  (154)  and  (109)  is  reduced  to 


tan'  (45°—  M)  =  ~  (104) 

tan  i  (c  +  a) 

which  will  give  a  more  accurate  result  than  (89),  when  A  is  nearly  90°. 
61.  From  (91)  we  find 

1  —  cos  b cos  a  —  cos  c 

1  -\-  cos  6      cos  a  -f-  cos  c 

or  tan*  £  b  =  tan  £  (c  +  «)  tan  £  (c  —  a)  (105) 

which  may  be  employed  instead  of  (91)  when  b  is  small,  or  nearly  180°. 
.  62.  From  (90)  we  find 

1  —  cos  B tan  c  —  tan  a 

1  -j"  cos  B      tan  c  -f-  tan  a 

tan1  £  B  =  sm  vc *v  (106) 

sin  (c  -j-  a) 

which  may  be  employed  instead  of  (90)  when  B  is  small,  or  nearly  180°. 
63.  By  similar  transformations  the  formulas  (101),  (102)  and  (103)  become 

tMl*e  =  ~c^i-~l*f)  (1°7) 

tan1  J  a  =  tan  [J  (A  +  5)  —  45°]  tan  [45°  +  J  U  —  B)l  (108) 

tan1  J  6  =  tan  [J  ( J  +  B)  —  45°]  tan  [45°  —  }  (A  —  B)]  (109) 

We  have  also,  by  (14), 

i  (A  -f-  J?)^  —  cos  (/ 


Bin*ie  =  —  „_.,.« 

2  sin  A  sin  /j 

which,  when  C  =  90°,  becomes 

"si  —  COS  (  s\   ~T~  -D I  / 1 1  /\\ 

Bin1  #  c  = • l  (11") 

2  sin  A  sin  B 

and  from  (15),  in  the  same  manner, 


2  sin  ^4  sin  P 


QUADRANTAL  AND  ISOSCELES  TRIANGLES.  177 

of  which  (110)  may  be  used  when  c  is  small,  and  (111)  when  c  is  nearly  180°,  instead 
of  (101). 

64.  The  equations  (92),  (93),  and  (94),  of  CASE  III.  give 

tan.  (45°  -  *  c)  =  ^"^7°}  (H2) 

tan  %  (A  +  a) 

tan'  (45°  -  *  b)  =  sjn  (A  ~  $  (113) 

sin  ( A  -f  a) 

tan1  (45°  —  $  B)  =  tan  i  (A  —  a)  tan  J  (A  +  a)  (114) 

The  roots  of  these  equations  having  the  double  sign,  we  may  take  the  angles 
450  —  J  c,  etc.  either  with  the  positive  or  negative  sign,  whence  the  two  solutions 
of  the  problem,  as  in  Art.  55. 

65.  Some  of  the  solutions  may  be  adapted  for  computation  by  the  table  of  natural 
sines.    Thus  from  (86),  (95),  and  (98), 

sin  a  =  i  [cos  (c  —  A)  —  cos  (c  +  ^4)]  (115) 

cos  B  =  *  [sin  (b  +  A)  —  sin  (b  —  A)]  (116) 

cos  c  =  J  [cos  (o  +  6)  -f-  cos  (o  —  &)]  (117) 

66.  The  following  relations  are  occasionally  useful : 
From  (83)  we  have 

cos  a sin  A  cos  A sin  2  A  f  118) 


cos  b  sin  B  cos  B      sin  2  -B 

From  (80)  and  (83), 

sin  J? sin  A  cos  A  __  sin  2  A 

sin  c  sin  a  cos  a        sin  2  a 

From  (80)  and  (84), 

sin.<4  sin  a  cos  a  sin  2  a 

cos  6  sin  c  cos  c        sin  2  c 


(119) 
(120) 


67.  Various  relations  may  be  deduced  from  the  general  formula  of  the  preceding 
chapter  by  making  C  =  90°.    The  following  are  easily  obtained  : 

sin  (c  —  a)  =  cos  c  tan  6  tan  \  B  —  cos  a  sin  b  tan  $  B 

sin  (c  -f  o)  =  cos  c  tan  b  cot  $  B  =  cos  a  sin  6  cot  J  B 

cos  (c  —  a)  =  cos  6  -\-  sin  a  sin  b  tan  $  B 

cos  (c  -{-  a)  =  cos  6  —  sin  a  sin  b  cot  $  B 

sin  (a  —  6)  =  2  sin  c  sin  $  (.4  -f  J5)  sin  $  (A  —  B) 

sin  (a  +  b)  =  2  sin  c  cos  J  (  J.  -f-  B)  cos  £  (^4  —  B) 

g}n  5  _  cosj_acosj_6        C03^>=       sin  }  a  sin  \b 


cos     c  cos    c 

Ian  S  =  —  cot  J  a  cot  J  b  (121) 

QUADRANTAL  AND  ISOSCELES  TRIANGLES. 

68.  The  polar  triangle  of  the  right  triangle  is  a  quadrantol  triangle,  one  side  (the 
side  opposite  the  angle  C)  being  equal  to  90°.  The  solution  of  such  triangles  is  as 
simple  as  that  of  right  triangles,  the  formulae  for  the  purpose  being  obtained  from  the 
preceding,  by  the  process  of  Art.  8.  It  is  unnecessary  to  produce  them  here,  as  quad- 
rantal  triangles  are  generally  avoided  in  practice,  and  when  unavoidable  are  readily 
solved  by  means  of  the  polar  triangle. 

An  isosceles  triangle  is  easily  solved  by  dividing  it  into  two  right  triangles  by  a  per- 
pendicular from  the  angle  included  by  the  equal  sides. 
23 


CHAPTER    III. 

SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES. 

69.  IN  the  solution  of  spherical  oblique  triangles,  a  required  part 
may  sometimes  be  found  by  its  sine,  in  which  case  there  will  be  two 
values  of  that  part,  answering  to  the  conditions,  unless  the  proper 
value  can  be  determined  by  other  considerations.     In  certain  cases, 
the  true  value  can  be  selected  by  applying  one  or  more  of  the  follow- 
ing principles,  some  of  which  are  demonstrated  in  geometry.  We  still 
consider  only  those  triangles  each  of  whose  parts  is  less  than  180°. 
I.   The  greater  side  is  opposite  the  greater  angle,  and  conversely. 
II.  Each  side  is  less  than  the  sum  of  the  other  two. 

III.  Tlie  sum  of  the  sides  is  less  than  360°. 

IV.  The  sum  of  the  angles  is  greater  than  1 80°. 

V.  Each  angle  is  greater  than  the  difference  between  180°  and  the 
sum  of  the  other  two  angles. 

For,  by  IV.,  A  +  B  +  C  >  180° 

whence,  A  >  180°  -  (B  +  C) 

But  if  B  +  C  >  180°,  we  have,  in  the  polar 
triangle,  A'B'C',  Fig.  8,  by  II., 

a'  <  V  +  c' 

or     180°  —  A<  180°  —  5+180°  —  C 
c,  -;1<1800-(J5+C) 

A>(B  +  C)-180° 

VI.  A  side  which  differs  more  from  90°  than  another  side,  is  in  the 
same  quadrant  as  its  opposite  angle. 

For,  by  (4),  we  have 

cos  a  —  cos  b  cos  c 
cos  A  =  — 

sin  6  sin  c 

in  which  the  denominator  is  always  positive.      If,  then,  a  differs 

178 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  179 

more  from  90°  than  b  or  than  c}  we  have,  (neglecting  the  signs  for  a 
moment), 

cos  a  >  cos  6  or  >  cos  c 

and  still  more  cos  a  >  cos  6  cos  c 

Hence  cos  a  being  numerically  greater  than  cos  b  cos  c,  the  sign  of  the 
whole  numerator,  and  therefore  the  sign  of  cos  A,  is  the  same  as  that 
of  cos  a;  that  is,  A  and  a  are  in  the  same  quadrant. 

VII.  An  angle  which  differs  more  from  90°  than  another  angle,  is 
in  the  same  quadrant  as  its  opposite  side.     For,  by  (5), 

cos  A  +  cos  B  cos  C 

cos  a  =  • 


sin  £  sin  C 

in  which,  if  A  differs  more  from  90°  than  B,  or  than  C,  cos  A  deter- 
mines the  sign  of  the  whole  fraction,  and  therefore  the  sign  of  cos  a. 

VIII.  In  every  spherical  triangle  there  are  at  least  two  sides  which 
are  in  the  same  quadrants  as  their  opposite  angles  respectively.     This 
follows  from  VI.  and  VII. 

IX.  The  sum  of  two  sides  is  greater  than,  equal  to,  or  less  than,  180°, 
according  as  the  sum  of  the  two  opposite  angles  is  greater  than,  equal 
to,  or  less  than,  180°.     In  other  words,  the  half  sum  of  two  sides  is 
in  the  same  quadrant  as  the  half  sum  of  the  opposite  angles.     For, 
by  (41), 

tan  \  (a  -f  6)  cos  J  (A  +  B)  —  tan  |  c  cos  %(A  —  B} 

the  second  member  of  which  is  always  positive,  so  that  tan  £  (a  +  6) 
and  cos  £  (-4  -f-  J5)  must  have  the  same  sign. 

70.  CASE  I.  Given  two  sides  and  the  in- 
cluded angle,  or  b,  c  and  A.  (Fig.  9.) 

First  Solution;  when  the  third  side  and 
one  of  the  remaining  angles  are  required. 

To  find  a.  The  relation  between  the  given 
parts  b,  c,  A  and  the  required  part  a  is  ex- 
pressed by  the  first  equation  of  (4), 

cos  a  =  cos  c  cos  b  +  sin  c  sin  b  cos  A  (M) 

by  which  a.  may  be  found  by  computing  separately  the  two  terms  of 
the  second  member  and  adding  their  values  to  form  the  natural 
cosine  of  a ;  but  we  should  thus  be  required  to  use,  besides  the  table 
of  log.  sines,  also  the  table  of  logarithms  of  numbers,  and  the  table 
of  natural  sines  and  cosines.  To  adapt  it  for  logarithmic  computa- 
tion by  the  table  of  log.  sines  exclusively,  we  employ  the  process  of 


) 

> 


180  SPHERICAL  TRIGONOMETRY. 

PI.  Trig.,  Arts.  174,  175.     Thus,  let  &  be  a  number  and  y>  an  aux- 
iliary angle  such  that 

k  sin  <p  =  sin  6  cos  A 

L 

K  COS  (p  =  COS  6 

then  (M)  becomes 

cos  a  =  k  (cos  c  cos  <p  -\-  sin  c  sin  y) 

=  k  cos  (c  —  <p)  (m') 

so  that  k  and  y>  being  found  from  (m)  we  may  find  a  by  (w').     But  we 
may  eliminate  k  by  dividing  the  first  equation  of  (m)  by  the  second, 

and  substituting  in  m'  the  value  of  k  =  -  —  ,  whence  we  have,  for 

cosy 
finding  a, 

tan  <p  =  tan  b  cos  .4 

(122) 


cos 


which  are  the  formulae  commonly  employed.* 

To  find  B.     The  relation  between  6,  c,  .4  and  J5,  is,  by  the  first 
equation  of  (10), 

sin  c  cot  b  —  cos  c  cos  A 


,  „ 

cot  B  = 


sin  A 


This  may  be  adapted  for  logarithms  by  the  process  above  em- 
ployed, but  to  assimilate  it  to  (M)  we  multiply  the  numerator  and 
denominator  of  the  second  member  by  sin  6,  whence 

,   n      sin  c  cos  b  —  cos  c  sin  6  cos  A 
cot  B  =  -        .        .      -- 
sin  b  sin  A 

which  by  (m)  becomes 

,,       k  sin  (c  —  y) 

cot  .Z?  =  —  (n) 

sm  6  sm  J 

or  substituting  the  value  of  k  =  -  -  ,  the  formulas  for  finding 

B  are  sin  V 

tan     =  tan  6  cos  A 

(123) 


sn  <p 


*  We  might  have  assumed  k  sin  <£  =  cos  ft,  A  cos  #  =  sin  b  cos  yl,  which  would  have 
reduced  (M)  to  COB  a  =  k  sin  (c  -}-  #).  In  this  way  all  the  solutions  that  follow  may 
be  varied. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  181 

In  the  use  of  these  formulae,  as  indeed  of  all  that  follow,  the  signs 
of  all  the  functions  must  be  carefully  observed,  according  to  PI.  Trig. 
Arts.  37  and  40. 

We  may  take  <p  between  0  and  180°,  less  or  greater  than  90°, 
according  as  the  sign  of  its  tangent  is  positive  or  negative;  or  we 
may  take  it  numerically  less  than  90°  in  all  cases,  but  positive  or 
negative  according  to  the  sign  of  its  tangent,  (PI.  Trig.  Arts.  37 
and  174). 

Check.     The  quotient  of  (ri)  divided  by  (HI')  is 

cot  B  _  tan  (c  —  tp) 
cos  a      sin  b  sin  A 

which  multiplied  by  the  following,  from  (3), 

sin  a  sin  B  =  sin  b  sin  A 
gives  tan  a  cos  B  —  tan  (c  —  y>)  (124) 

by  which  the  values  of  a  and  B,  found  by  (122)  and  (123),  may  be 
verified. 

71.  If  a  and  C  were  required,  the  solution  would  evidently  be 
similar,  only  interchanging  6  and  c,  B  and  C.  By  the  fundamental 
formulae  we  should  have 

cos  «  =  cos  6  cos  c  -\-  sin  6  sin  c  cos  A 

n      sin  b  cos  c  —  cos  b  sin  c  cos  A  f     (O) 

cot  o  —  -  - 

sin  c  sin  A 

and  denoting  the  auxiliary  angle  in  this  case  by  £,  the  logarithmic 
solution  would  be 

tan  /  —  tan  c  cos  A 

cos  (b  —  y]  cos  c 
cos  a  =  --  >  -  L1— 
cosy 

025) 


sin/ 

Check,     tan  a  cos  C  =  tan  (6  —  /) 

Q 


182  SPHERICAL  TRIGONOMETRY. 

EXAMPLES. 

1.  Given  6  =  120°  30'  30",    c  =  70°  20'  20",    A  =-  50°  10'  10"; 
find  a  and  B. 

By  (122). 

6  =       120°  30'  30"        log  tan  b  —  0-2297071 
A  =        50°  10'  10"        log  cos  A  -f  9-8065322 
y  =        132°  36'  44"-2*  log  tan  y  —  0-0362393 
c  =        70°  20'  20"-0 
c  —  y  =  --    62°  16'  24"-2 

By  (122).  By  (123).  By  (124). 

log  cos  (e  —  0)  +  9-6676893  log  sin  (c  —  <j>)  —  9'9470304  log  tan  (c  —  <t>)  —  0'2793410 

ar  co  log  cos  ^  —  0-1693898  ar  co  log  sin  <j>  -f  G'1331505  log  tan  a  +  0-4291648 

log  cos  6  —  9-7055761  log  cot  A  +  9'9212038  log  cos  B  —  9-8501762 

log  cos  a  +  9-5426552  log  cot  B  —  0'0013847  Check.     —  0'2793410 

a  =  69°  34'  55"-9  B  =  135°  5'  28"'8 

2.  Given  b  =  120°  30'  30",    c  =  70°  20'  20",    A  =  50°  10'  10"; 
find  a  and  C. 

Ans.     a  =  69°  34'  55"-9 
C=50°30'    8"-4 

3.  Given  b  =  99°  40'  48",  c  =  100°  49'  30",  A  =  65°  33'  10"; 
find  a  and  B. 

Ans.  a  =  64°  23'  15"'0 
B  =  95°  38'  4"-0 

4.  Given  b  =  99°  40'  48",   c  =  100°  49'  30",    A  =  65°  33'  10"; 
find  a  and  O. 

Ans.     a  =  64°  23'  15"-0 
C=  97°  26'  29"-l 

5.  Given  b  =  98°  2'  20",   c  =  80°  35'  40",   A  =  10°  16'  30" ; 
find  a  and  C. 

Ans.  a  =  20°  13'  30"'l 
C=  30°  35'  56"-7 

72.  If  B,  C  and  a  were  all  required,  we  might  find  a  and  C  by 
(125),  and  then  B  by  Art.  3,  which  gives 

sin  a  :  sin  b  =  sin  A  :  sin  1? 
sin  b  sin  .4 


or  sin  .B  = 


sin  a 


*  We  may  also  take  ^  =  —  47°  23X  15"'8,  whence  c  —  <j>  =  117°  43'  35"'8.  which 
will  give  the  same  values  of  a  and  B  as  found  in  the  text. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  183 

Of  the  two  values  of  B  less  than  180°  given  by  this  formula,  the 
proper  one  may  generally  be  selected  by  the  principles  of  Art.  69. 
There  are  cases,  however,  in  which  all  the  conditions  there  given  are 
satisfied  by  both  values  of  B,*  and  on  this  account  it  is  preferable, 
in  general,  to  combine  (123)  and  (125),  or  to  employ  the  following 
solution,  when  the  three  unknown  parts  are  all  to  be  found. 

73.  CASE  I.  Given  6,  c  and  A.  Second  Solution;  when  the  two 
remaining  angles  are  required,  or  when  the  three  unknown  parts  are 
all  required. 

We  have,  by  Napier's  Analogies,  (42)  and  (43), 

sin  £  (b  +  c)  :  sin  \  (b  —  c)  =  cot  \A  :  tan  %  (B  —  C) 
cos  |  (6  +  c)  :  cos  £  (6  —  c)  =  cot  £  A  :  tan  £  (B  +  C) 
whence 

tan  K*  -  C7)  =  S!"~1  cot  *  A  (126) 


sin 


tan  I  (B  +  C)  =  ^-  Cot  *  A 


COS 


which  determine  J  (B  —  (?)  and  J  (B  +  C)  ;  then  the  half  difference 
added  to  the  half  sum  gives  the  greater  angle,  and  the  half  difference 
subtracted  from  the  half  sum  gives  the  less  angle. 

If  c  >  6,  we  may  write  c  —  6,  C  —  B,  in  the  place  of  6  —  c,  B  —  C. 

We  may  now  find  a  by  either  of  Napier's  Analogies,  (40),  (41), 
which  give  f 

4          <128> 

'>      (129> 


*  By  Art.  69,  VI.,  if  6  differs  more  from  90°  than  c,  B  is  in  the  same  quadrant  as  b, 
and  all  ambiguity  is  removed.  If  c  differs  more  from  90°  than  6.  we  may  find  a  and  B 
by  (122)  and  (123),  and  then  C  by  the  formula 

sin  c  sin  A 
sin  C— 


sin  a 

C  being  taken  in  the  same  quadrant  as  c. 

f  We  may  also  find  a  from  any  one  of  Gauss's  Equations  (44),  which  become,  in  th* 
present  case, 

cos  J  a  sin  J  (B  -f-  C)  =  cos  J  A  cos  \  (6  —  c) 
cos  \  a  cos  £  ( B  4-  (7)  =  sin  J  A  cos  £  (6  -f-  c) 
sin  £  a  sin  J  (5  —  C)  =  cos  J  ,4  sin  J  (6  —  c) 
sin  }  a  cos  }  (J5  —  (7)  =  sin  J  .4  sin  £  (6  -{-  c) 


184 


SPHERICAL  TRIGONOMETRY. 


EXAMPLES. 

1.  Given  6  =  120°  30'  30",  c  =  70°  20'  20", 
find  J5,  C  and  a.     We  have 

£  (&  +  c)  =  95°  25'  25" 

i  tfo c\  _ _  25°    5'    5" 

£4  =  25°    5'    5" 
By  (126). 

ar  co  log  sin  £  (6  +  c)  +  0-0019487 
log  sin  %  (b  —  c)  +  9-6273228 
log  cot  £  4  +  0-3296529 
log  tan  |(5  —  C)  +  9-9589244 
—  <7)=   42°  17'40"-2 


=  50°  10'  10"; 


By  (127). 

ar  co  log  cos  £  (6  +  c)  —  1-0244829 
log  cos  %  (b  —  c)  +  9-9569757 
log  cot  £  A  +  0-3296529 


5  =  135°    5' 28"-8 

By  (128). 

ar  co  log  sin  ^(5—  C)+0«l  720227 

log  sin  £  (JS+  C)+9-9994824 

log  tan  £  (6  —  c)+9-6703471 

log  tan  £a+9-841 8522 

a  =  34°47'28"-0 


log  tan  ^  (5  -f  (7)  —  1-3111115 
£(£  +  <?)  =  92°  47'48"-6 
C=  50°  30'    8"-4 

By  (129). 

ar  co  log  cos  $(B—  C)-fO-1309469 

log  cos  %(B+  C)—  8-6883709 

log  tan  \  (b  +  c)—  1-0226342 

logtan£a+9-8418520 


5'  28"-8 
30'  8"-4 
34'56"-0 


C=    50 
a=   69 


2.  Given  6  =  99°  40' 
find  JS.  C  and  a. 


48",  c=  100°  49'  30",  A  =  65°  33'  10"; 

Ans,  5  =  95°  38'  4"-0 
C=97°  26'  29"-l 
a  =  64°  23'  15"-1 

74.  It  may  be  remarked  with  regard  to  (128)  and  (129)  that,  when  b  and  c  (and 
consequently  B  and  C)  are  nearly  equal,  a  small  error  in  the  previous  determination 
of  the  small  angle  J  (JS  —  C)  may  produce  a  large  one  in  log  sin  £  (B—  C),  and 
consequently  in  log  tan  J  a  found  by  (128).  In  that  case,  therefore,  (129)  must  be 
preferred. 

In  like  manner,  if  $  (b  +  c),  and  consequently  J  (B  +  C),  are  nearly  equal  to  90°, 
(129)  will  become  inaccurate,  and  then  (128)  is  to  be  preferred. 

Formula  (128)  would  fail  entirely  if  B  =  C,  and  formula  (129)  would  fail  if 
J  (B  -f  C)  —  90°,  since  the  second  members  in  these  cases  would  assume  the  inde- 
terminate form  — 

75.  CASE  I.  Given  6,  c  and  A.  Third  Solution.  When  the 
third  side  is  alone  required,  the  computation  by  (122)  is  in  most  cases 
as  convenient  as  any  other;  but  there  are  various  other  methods 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  185 

derived  from  the  formulae  of  the  preceding  chapter,  which  have  been 
employed  with  advantage  in  particular  applications.  Among  the 
most  convenient  are  the  following,  from  (12)  and  (13): 

cos  a  =  cos  (6  —  c)  —  2  sin  6  siu  c  sin2 1-  A  (130) 

cos  a  =  cos  (6  +  c)  +  2  sin  6  sin  c  cos2  £  A  (131) 

The  computation  of  these  requires  the  use  of  natural  cosines  and 
numbers,  the  signs  of  which  must  be  carefully  observed. 

EXAMPLE. 

Given  6  =  99°  40'  48",  c  =  100°  49'  30",  A  =  65°  33'  10"; 
find  a. 

By  (130).* 

£  A  =      32°  46'  35"         log  sin2  $  A  =  2  log  sin  £  A     9*4669752 
6  — c=—    1°    8' 42"  log  sine    9-9922023 

log  sin  6     9-9937722 
log  2    0-3010300 

-  2  sin  b  sin  c  sin8  £  A  =  —  0-5675181  log     9-7539797 

nat  cos  (6  —  c)  =  +  0*9998003 

nat  cos  a  =  +  0-4322822  a  =  64°  23'  15" 

By  (131). 

$A=   32°  46' 35"  log  cos2  £  A  =  2  log  cos  $  A  9-8493748 

6  -f  c  =  200°  30'  18"  log  sin  c  9-9922023 

log  sin  6  9-9937722 

log  2  0-3010300 

-f  2  sin  b  sin  c  cos2  %A=  +  1-3689240  log  0-1363793 

nat  cos  (6  +  c)  =  —  Q-9366416 

nat  cos  a  =  +  0-4322824  a  =  64°  23'  15" 

76.  In  Art.  14,  we  have  deduced  several  formulae  by  which  J  a  may  be  computed. 
We  may  adapt  (17)  and  (18)  for  logarithmic  computation,  as  follows: 

sin*  0  =  sin  6  sin  c  cos*  J  A 
sin*  J  a  =  sin*  J  (6  +  c)  —  sin*  ^  (132) 

=  sin  [J  (6  +  c)  +  0]  sin  [J  (6  +  c)  -0] 

sin*  *  =  sin  b  sin  c  sin*  J  .4 
cos*  Ja  =  cos*  J  (6  — c)— sin*0  (133) 

=  cos  [*  (6  -  c)  +  f|  cos  [H*  -  <0  -  fl 
of  which  (132)  is  to  be  preferred  when  J  a  <  45°,  and  (133)  when  £  a  >  46°. 

*  The  computation  of  (130)  is  facilitated  by  the  use  of  a  special  table  (given  in  many 
treatises  on  navigation),  from  which,  with  the  argument  A,  is  taken  the  logarithm  of 
2  sin*  J  A  =  versin  A.     [PI.  Trig.  (4)  and  (139)]. 
24  <j2 


186  SPHERICAL  TRIGONOMETRY. 

77.  CASE  II.  Given  two  angles  and  the 
included  side,  or  A}  C  and  b.  (Fig.  9). 

First  Solution;  when  the  third  angle  and 
one  of  the  remaining  sides  are  required. 

To  find  B.  The  relation  between  A,  C,  h 
and  .B,  is,  by  (5), 

cos  B  =  —  cos  C  cos  A  -\-  sin  C  sin  A  cos  b  (M) 

which  is  adapted  for  logarithms  by  the  method  employed  in  the  pre- 
ceding case.     Thus,  let 

h  sin  #  =  cos  A 

V        (m) 
h  cos  •&  =  sin  A  cos  b  J 

then  (M)  becomes 

cos  S  =  h  (sin  C  cos  $  —  cos  C  sin  $) 


f*rm  ^ 
or,  eliminating  A  =    .        *  the  formulae  for  finding  B  are 

cot  d  =  tan  -4  cos  b 

sin  (C-  #)  cos  A  (134) 

cos  5  =  --  L—  r-^- 

sm# 

a.     From  the  third  equation  of  (10),  we  find, 
sin  C  cot  A  +  cos  C  cos  b 


cot  a  = 

sin  b 

esi  n   ('  /•»r\c»      ^         '       r»*~vC?    f    '  C3i  fk      ^J     OOS   O 

(N) 


sin  A  sin  6 
which,  by  (m),  becomes 

hcos(C—  #) 

cot  a  =  -  (n) 

sin  yl  sin  6 

,.    .     ,.       ,       sin  A  cos  6  r     ,,    ,. 

or.  eliminating  h  =  —  —  >  we  have,  tor  finding  «, 

cos  v 

cot  ^  =  tan  A  cos  6 

cos  (C—  #)  cot  6  }•      (135) 

cot  a  =  - 


As  in  the  preceding  case,  we  may  either  take  $  always  between  0 
and  180°,  less  or  greater  than  90°  according  as  its  tangent  is  posi- 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  187 

tive  or  negative;  or  we  may  take  &  numerically  less  than  90°  in  all 
cases,  positive  or  negative,  according  to  the  sign  of  its  tangent. 
(PI.  Trig.  Art.  174.) 

Check.  The  quotient  of  (n)  by  (w')  is 

cot  a  _  cot  (C  —  #) 
cos  B        sin  A  sin  6 
which,  multiplied  by 

sin  £  sin  a  =  sin  A  sin  b 
gives 

tan  B  cos  a  =  cot  (C—  #)  (136) 

by  which  the  values  of  B  and  a,  found  by  (134)  and  (135),  may  be 
verified. 

78.  If  B  and  c  were  required,  the  solution  would  be  similar,  only 
interchanging  a  and  c,  A  and  C.  By  the  fundamental  formulae,  we 
should  have, 

cos  B  =  —  cos  A  cos  (7+  sin  A  sin  Ccos  b 

sin  A  cos  (7+  cos  A  sin  C  cos  6 
cot  c  =  -      -  r-        .      -- 
sin  C  sin  6 

and  denoting  the  auxiliary  angle  by  £,  the  logarithmic  solution 
would  be 

cot  £  =  tan  C  cos  6 


„       sin  (A  —  f)  cos  C 

B  =  — 

sin  r 

cos  (-4  —  C)  cot  6 
-  -^  --- 


(137) 

n.ns  i  A  —  r  \  r>nr.  n 

cot  c 

COS  £ 

Check,     tan  B  cos  c  =  cot  ( A  —  £) 

EXAMPLES. 

1.  Given  A  =  135°  5'  28"-8,  C=  50°  30'  8"-4,  6  =  69°  34'  55"-9; 
find  B  and  a. 

By  (134). 

A  =      135°    5'  28"-8  log  tan  ^1  —  9-9986154 

6  =        69°  34'  55"-9  log  cos  b  -f  9-5426553 

tf  =      109°  10'  31"-0*          log  cot  d  —  9-5412707 
C=        50°  30'    8"-4 
C—&  =  -     58°40'22"-6 

*  We  may  also  take  #  =  —  70°  49'  29/A0,  whence  (7—  *  =  121°  19r  37/A4,  which 
will  evidently  give  the  same  results  as  those  obtained  in  the  text. 


188  SPHERICAL  TRIGONOMETRY. 

By  (134).  By  (135).  By  (136). 

log  sin  (C—  tf)  —  9-9315664  log  cos  ((?—«)  +  9'7159386  log  cot  (C—  *)  —  9'7843722 

ar  co  log  sin  $  +  0*0247897  ar  co  log  cos  #  —  0-4835187            log  tan  B  +  0-0787962 

log  cos  A  —  9-8501762  log  cot  6  -f  9'5708352            log  cos  a  —  9*7055757 

log  COB  £  +  9-8065323  log  cot  a  —  9'7702925               Check.    —  9'7843719 

B  =  50°  10'  10"'0  a  =  120030'29"-9 

Ans.  B  =    50°  10'  10"-0 


2.  Givenjt  =  135°5'28"-8,  (7=50°  30'  8"-4,  b  =  69°  34'  55"-9; 
find  P>  and  c. 

Ana.  B  =    50°  10'  10"-0 
c=    70°20'20"-0 

3.  Given  A  —  65°  33'  10",    C=  95°  38'  4",    b  =  100°  49'  30"  ; 
find  B  and  a. 

Am.  B  =    97°  26'  29" 
o=    64°  23'  15" 

4.  Given  4  =  97°  26'  29"       (7=  95°  38'  4",     6  =  64°  23'  15"  ; 
find  B  and  a. 

4n*.  B  =    65°  33'  10" 
a  =  100°  49'  30" 

79.  If  a,  c  and  J2  were  all  required,  we  might  find  B  and  c  by 
(137),  and  then  a  by  Art.  3,  which  gives, 

sin  B  :  sin  A  =  sin  b  :  sin  a 

sin  ^4  sin  6  7       . 

sin  a  —  --  —  (138) 

sm  .B 

Of  the  two  values  of  a  given  by  this  equation,  the  proper  one  is  to 
be  selected,  if  possible,  by  the  principles  of  Art.  69.*  But  as  cases 
occur  in  which  all  the  conditions  there  given  are  satisfied  by  both 
values  of  a,  it  is  preferable,  in  general,  to  combine  (135)  and  (137), 
or  to  employ  the  following  solution  when  the  three  unknown  parts 
are  all  to  be  found. 


*By  Art.  69,  VII.,  when  A  differs  more  from  90°  than  C,  a  must  be  taken  in  the 
same  quadrant  with  A,  and  all  ambiguity  is  removed.  If,  then,  by  A  we  always 
denote  that  angle  which  differs  more  from  90°  than  the  other  given  angle,  we  may 
always  solve  this  case  by  means  of  (137)  and  (138),  without  meeting  with  any  difficulty 
in  determining  the  quadrant  in  which  a  is  to  be  taken. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TKIANGLES.  189 

80.  CASE  II.  Given  A,  C  and  6.     Second  Solution;  when  the  two 
remaining  sides,  or  when  the  three  unknown  parts  are  all  required. 
We  have,  by  Napier's  Analogies,  (40)  and  (41), 

sin  •£  (A  +  (7)  :  sin  \  (A  —  C)  =  tan  -|  b  :  tan  |-  (a  —  c) 
cos  £  (A  +  C)  :  cos  £  (A  —  C)  =  tan  1  6  :  tan  $  (a  +  c) 
whence 

—  C)         j  , 


(139) 
** 


which  determine  -^  (a  —  c)  and  ^  (a  +  c)  ;  then  the  half  difference 
added  to  the  half  sum  gives  the  greater  side,  and  the  half  difference 
subtracted  from  the  half  sum  gives  the  less  side.  If  C>A,  we  may 
write  C  —  A,  c  —  a  in  the  place  of  A  —  (7,  a  —  c. 

We  may  now  find  B  by  either  of  Napier's  Analogies,  (42)  and 
(43),  which  give* 


sn 


(a  -f  c)         .  ,  .       ^x  ,.,  A  . 

J  -  (  tan  |(  A  —  0)  (140) 


sin  •£  (a  —  c) 

tan  %(A+C)  (141) 


cos  •£  (a  —  c) 

EXAMPLES. 

1.  Given  J.  —  135°  5'  28"-6,  C=  50°  30'  8"-6,  6  =  69°  34'  56'r'2  ; 
find  a,  c  and  J?. 

We  have  A+C  =  92°  47'  48"'6 


^  6  =  34°  47'  28"-l 
Then,  by  (139), 

ar  co  log  sin  ±(A+  C)4-0'0005176     ar  cologcos|(^4+C)-l'3116286 

log  sin  %(A—  C)+  9-8279768  log  cos  $  (A—  C)+9'8690535 

log  tan  |  b  +  9-841  8527  log  tan  ^6  +  9-8418527 


log  tan  $  (a—  c)  +  9-6703471  log  tan  |(a  +  c)  —  1-0225348 

%(a  —  c)=    25°    5'    5"-0  ^  (a  +  c)  =  95°  25'  25"'0 

a  =  1  20°  30'  30"-0  c  =  70°  20'  20"-0 


*  We  may  also  find  B  by  any  one  of  Gauss's  Equations,  (44),  interchanging  B  and 
C,  b  and  c. 


190  SPHERICAL  TRIGONOMETRY. 

By  (140).  By  (141). 

ar  co  log  sin  J  (o  —  c)  +  0-3726772  ar  co  log  cos  J  (a  —  c)  +  0-0430243 

log  sin  J  (a  +  c)  +  9'9980523  log  cos  J  (a  +  c)  —  8-9755171 

log  tan  I  (A—  C)+  9-9589234  log  tan  $(A+C)  — 1-3111110 

log  cot  £  B  +  0-3296529  *  log  cot  |  £  +  0-3296524 

£  5  =  25°  5'  5"-0 

4n*.      a  =  120°  30' 30" 

c  =    70°  20'  20" 

B  =    50°  10'  10" 

2.  Given 4  ==  95°  38'  4",  C=  97°  26'  29",  6  —  64°  23'  15"; 
find  a,  c  and  B. 

Ans.   a=  99°  40' 48" 

c  =  100°  49'  30" 

B  =  65°  33'  10" 

81.  CASE  II.  Given  A,  C  and  b.  Third  Solution.  When  the 
third  angle  B  is  alone  required,  the  computation  by  (134)  is  in  most 
cases  as  convenient  as  any  other,  but  there  are  other  methods  (cor- 
responding to  those  given  in  Art.  75  for  finding  a)  which  may  occa- 
sionally be  serviceable.  By  (14)  and  (15)  we  have 

cos  B  —  —  cos  (A  -+-  C)  —  2  sin  A  sin  C  sin2  £  6  (142) 

cos  B  =  -  cos  (A  —  C)  +  2  sin  A  sin  Ccos2  ^6          (1 43) 

the  computation  of  which  is  similar  to  that  of  (130)  and  (131). 

EXAMPLE. 

Given  A  =  95°  38'  4",  C  —  97°  26'  29",  6  =  64°  23'  15" ; 
find  £. 

By  (142). 

£6=    32°ll'37"-5         log  sin2 ^6  =  2  log  sin  16  9-4531022 

A  -f  C=  193°    4'  33"                                        log  sin  A  9-9978967 

log  sin  C  9-9963268 

log  2  0-3010300 

-  2  sin  A  sin  C  sin2  £  6  =    -  0-56021 62  log     9-7483557 

-  nat  cos  (A  +  C)  ==  +  Q'9740715 

nat  cos  B  =  +  0.4138553  B  —  65°  33'  9"'9 

*For  the  reasons  given  in  Art.  74,  (141)  is,  in  this  example,  not  so  accurate 
as  (140). 


SOLUTION  OF  SPHERICAL  OBLIQUE  TKIAISGLES. 


191 


82.  In  Art.  14,  several  formulae  are  given,  by  which  £  B  way  be  computed.    By  (21) 
and  (22)  we  have 

sin2  \  B  =  cos2  J  (A  —  C)  —  sin  A  sin  C  cos2  £  6 
cos1  £  B  =-  sin2  £  (A  +  C)  —  sin  A  sin  C  sin2  J  6 

which  may  be  adapted  for  logarithms,  thus  : 

sin2  <]>  =  sin  A  sin  C  cos2  £  6  "j 

sin*  £  £  =  cos2  J  (  J.  —  (7)  —  sin2  <j>  I     (144) 


sin1  #  =  sin  A  sin  (7  sin2  J  6 
cos2  IB  =  sin2  \  (A  +  C)  —  sin1  0 
=  sin  [J  (4  +  C)  +  *]  sin 


(145) 


of  which  (144)  is  to  be  preferred  when 


+  (7)  -  0] 
<  45°,  and  (145)  when  J  J?  >  45°. 


83.  Case  II.  might  have  been  reduced  to  Case  I.  by  means  of  the  polar  triangle, 
Art.  8 ;  for  there  will  be  known  in  the  polar  triangle,  two  sides  and  an  angle  oppo- 
site one  of  them,  being  the  supplements  of  the  given  angles  and  side  of  the  pro- 
posed triangle.  The  polar  triangle  being  solved,  therefore,  by  Case  I.,  and  its  two 
remaining  angles  and  third  side  found,  the  supplements  of  these  parts  would  be  the 
two  sides  and  third  angle  required  in  the  proposed  triangle.  It  is  easily  seen,  also, 
that  all  the  formulae  above  given  for  this  case  might  have  been  obtained  by  these 
considerations. 

84.  CASE  III.     Given  two  sides  and  an  FlG-9-    o 

angle  opposite  one  of  them;  or  a,  6,  and  A. 
Fig.  9. 

First  Solution,  in  which  each  required  part 
is  deduced  directly  from  fundamental  for- 
mulae independently  of  the  other  two  parts. 

To  find  c.     We  have,  by  (4),* 

cos  c  cos  6  -j-  sin  c  sin  6  cos  A  =  cos  a 
to  solve  which,  let 

k  sin  <p  =  sin  b  cos  A 

k  cos  <p  —  cos  6 
then  (M)  becomes 

k  cos  (c  —  <p)  =  cos  a 
or  putting  c  —  <f>  —  (p1, 

k  cos  (p1  =  cos  a  ) 

c  =  <f>  -f  <p'  ( 

*  This  formula  has  been  already  employed  and  adapted  for  logarithms  in  Case  1  ; 
but,  for  the  sake  of  clearness,  it  is  repeated.  The  student  will  remark  that  a  sim- 
ple transformation  of  (122)  gives  (146).  It  will  also  be  observed  that  the  given  angle 
and  the  given  side  adjacent  to  it,  in  each  of  the  first  four  cases,  are  denoted  by  A 
and  b,  in  order  that  the  auxiliaries  <j>  and  i?  may  have  the  same  values  throughout. 


(M) 


(m) 


192  SPHERICAL  TRIGONOMETRY. 

The  auxiliary  <p  will  be  fully  determined  by  (m),  being  taken  be- 
tween 0  and  180°,  and  always  positive  (PI.  Trig.  Art.  174);  but,  as 
the  cosine  of  an  angle  is  also  the  cosine  of  the  negative  of  that  angle 
[PI.  Trig.  (56)],  we  may  take  y>f  in  (m'}  either  with  the  positive  or 
the  negative  sign,  so  that  c  =  y  ±  <p'.  There  will  thus  be  two  values 
of  c  answering  to  the  same  data,  both  of  which  will  be  admissible, 
except  when  <p  -\-  (pr  exceeds  180°,  in  which  case  the  only  solution 
is  c  =  <p  —  <pf  •  and  except  when  <p'  exceeds  <p  (which  would  make 
c  negative),  in  which  case  the  only  solution  is  c  =  <p  +  <p'. 

Therefore,  eliminating  k,  we  have  for  finding  c, 

tan  <p  —  tan  b  cos  A 

,      cos  <p  cos  a 

cos  <p'  = T- 

cos  6 

C  =  (D  ~*~  ID1 

To  find  C.    We  have  by  (10), 

cos  C  cos  b  -f-  sin  C  cot  A  =  sin  b  cot  a 
or,  multiplying  by  sin  A, 

cos  C  sin  A  cos  6  -f  sin  C  cos  A  =  sin  A  sin  6  cot  a          (N) 
to  solve  which,  let 

A  sin  $  =  cos  A 

V         (n) 
h  cos  &  =  sin  A  cos  6  J 

then  (N)  becomes 

h  cos  (C —  $)  =  sin  A  sin  b  cot  a 
or  putting  {?—#  =  #', 

h  cos  #'  =  sin  A  sin  b  cot  a  | 

_  ,  M 

Here  $  will  be  fully  determined,  while  &'  found  by  its  cosine  may 
be  either  positive  or  negative,  so  that  we  shall  have  in  general  two 
values  of  C—&  ±&f,  corresponding  respectively  to  the  two  values 
of  c;  but,  as  before,  values  greater  than  180°,  and  negative  values, 
being  excluded,  there  will  in  certain  cases  be  but  one  solution. 

.       sin  A  cos  b  f     »    ,.       ^ 

Eliminating  h  =  -  — >  we  have,  then,  for  finding  C} 

cos  v 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  193 

cot  &  =  tan  A  cos  6  ^ 

cos  ft'  —  cos  &  tan  b  cot  a  I      (147) 

c=&±#> 

To  find  B.     We  have  several  methods:  1st,  directly  by  (3), 

.     D       sin  A  sin  6  ,       . 

sin  B  —  -  (148 ) 

sin  a 

which  gives  two  values  of  B,  supplements  of  each  other,  correspond- 
ing respectively  to  the  two  values  of  c  and  C.  We  shall  presently 
see  how  to  determine  which  are  the  corresponding  values  of  c, 
C  and  B. 

2d.  In  (123),  <p  has  the  same  value  as  in  (146),  and  therefore  put- 
ting in  (123),  c  —  <p  =  tp',  we  have 

,   n      sin  <p'  cot  A 
cot  B  — * (149) 

sm  <p 

which  gives  two  values  of  B  by  the  positive  and  negative  values 
of  <p'. 

3d.  By  (124), 

cos  B  =  tan  <pf  cot  a  (1 50) 

which  also  gives  two  values  of  B  by  the  positive  and  negative  values 
of  (f1. 

4th.  In  (134),  $  has  the  same  value  as  in  (147),  and  therefore  put- 
ting in  (134),  C—ft  =  ft', 

D       sin  ft'  cos  A 

cos  B  = (151) 

sin  ft 

which  gives  two  values  of  B,  as  before. 
5th.  By  (136), 

cot  B  =  tan  #'  cos  a  (152) 

which  gives  two  values  of  B,  as  before. 

The  formula  (149)  shows  that  when  <p'  is  positive,  cot  B  and  cot  A 
have  the  same  sign,  that  is,  B  and  A  are  in  the  same  quadrant ;  and 
that,  when  <p'  is  negative,  cot  B  and  cot  A  have  different  signs,  that 
is,  B  and  A  are  in  different  quadrants.  A  like  result  follows  from 
(151),  with  reference  to  $'.  Hence,  that  value  of  B  which  is  in  the 
same  quadrant  as  A,  belongs  to  the  triangle  in  which  c  =  <p  +  <p', 
C=  &  -f  ft' ;  and  that  value  of  B  which  is  in  a  different  quadrant 
from  A,  belongs  to  the  triangle  in  which  c  =  y  —  <pf,  C—ft  —  ft'. 
This  precept  enables  us  to  employ  (148)  without  ambiguity.  In  the 

25  R 


194  SPHERICAL  TRIGONOMETRY. 

use  of  (149),  (150),  (151)  and  (152),  it  is  only  necessary  carefully  to 
observe  the  signs  of  the  several  terms. 

Checks.  Of  the  various  formulae  above  given  for  finding  B,  one  or 
more  may  be  employed  for  the  purpose  of  verification.  When  c  and  C 
have  been  found,  the  most  simple  check  is  the  following,  from  (3), 

sin  C  _  sin  4 
sin  c        sin  a 

which,  indeed,  might  have  been   employed  to  find   (7,  after  c  was 
found,  and  reciprocally,  but  for  the  ambiguity  attaching  to  the  sines. 

85.  According  to  Art.  69,  VI.,  if  b  differs  more  from  90°  than  a, 
B  must  be  in  the  same  quadrant  as  6,  and,  since  but  one  of  the  two 
values  of  B  can  satisfy  this  condition,  there  will  be  but  one  solution. 
In  that  case  c  and  C  will  each  be  found  to  have  but  one  admissible 
value. 

86.  The  problem  will  be  altogether  impossible,  when  a  differs  more 
from  90°  than  6,  and  is  yet  not  in  the  same  quadrant  with  A.     In 
such  case,  we   should    find  that   tp  +  <p'  >  180°,  and    <p  —  <pf  <  0; 


The  problem  will  also  be  impossible,  when  sin  A  sin  6  >  sin  a, 
since,  by  (148),  we  shall  then  have  sin  B  >•  1. 

EXAMPLES. 

1.  Given  a  =  40°  16',  6  =  47°  44',  A  =  52°  30';  find  B. 

By  (148). 

a=    40°  16'  ar  co  log  sin   a  0*1895350 

b  =    47°  44'  log  sin  6    9-8692449 

A  =    52°  30'  log  sin  A  9-8994667 

B=    65°  16'  35"  log  sin  B  9-9582466 

or         7?  ==11  4°  43'  25" 

2.  With  the  same  data,  find  c  and  B. 

By  (146). 

a=       40°  16'  logcos  a+9-8825499 

b=        47°  44'         log  tan  6+0-0414996  ar  co  log  cos  6+0-1722547 
A  =        52°  30'         log  cos  .4+9-7844471 

tp=       33°  48'  51  "-4  logtan^p+  9-8259467          log  cos  ^+9-9195204 

<p'  =  ±   19°30'29"-0  log  cos  <p'+  9-9743250 

c,=        53°19'20"-4 
c  =       14°  18'  22"-4 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  195 

By  (149).  Check.  (150). 

tf,  =  33°  48'  51  "'4  ar  co  log  sin  ^+0-2545328  log  cot  «+0-0720848 
y*  =±19°  30'  29"-0  log  sin  ^'±9-5236676  log  tan  ^'±9-5493427 
A=  52°  30'  0"  log  cot  ^+9-8849805  ±9-6214275 

Bl  -      65°  16'  £  ,  £±9-6631809  logcos£±9-6214275 

£2  =    114°43'25"-1J 

Ana.  c  =  53°  19'  20"-4  )  f  c  =     14°  18'  22"-4 

B  =  65°  16'  34"-9  j          j 5  =  114°  43'  25"-l 

3.  Given  a  =  120°,  b  =  70°,  .4  =  130° ;  find  C  and  B. 

By  (147). 

a=      120°  log  cot  a— 9-7614394 

b  =       70°  log  cos  b  +  9-5340517  log  tan  6  +0-4389341 

A=     130°  log  tan  ^—0-0761865 

#=  112°10'33"-6  log  cot  #  —  9-6102382  log  cos  #—9-5768627 
*'=±  53°13'13"-8  log  cos  #'+9-7772362 

Q=     165°  23'  47"-4 
C2=       58°  57'  19"-8 

By  (151).  CAecfc.  (152). 

*=  112°10'33"-6  ar  co  log  sin  #+0-0333755  log  cos  a  —9-6989700 
#'=±  53°13'13"-8  log  sin  #'±9-9036030  log  tan#' ±0-1 263669 

^4=    130°    0'    0"  log cos4—9-8080675  +9-8253369 

7?==    19'}°  4fi'  ^T7'- 


log  cos  B+  9.7450460  log  cot  B+  9-8253369 
=     56°13'22"-5' 

•  Ana.     C=165°23'47"-4j    ^    (  C=  58°  57'  19"-8 
B  =  123°46'37"-5j          \B  =  56°  13'  22"-5 

4.  Given  a  =  70°,  b  =  120°,  A  =  130°;  find  C. 

By  (147). 

a  =       70°  log  cot  a  +  9*5610659 

b=     120°  log  cos  6  —  9-6989700  log  tan  6  —  0-2385606 

A=     130°  log  tan  A—  0-0761865 

#=        59°12'37"-0     log  cot  #+  9-7751565  log  cos  #+  9-7091756 
&'=±  108°  49'  35"- 1  log  cos  #'—9-5088021 

C=      168°    2'  12"-1,  taking  #'  with  the  positive  sign  only,  since  its 
negative  value  would  render  0  negative. 


196 


SPHERICAL  TRIGONOMETRY. 


5.  Given  a  =  99°  40'  48",  b  ==  64°  23'  15",  A  =  95°  38'  4" ;  find 
c,  Cand  J?. 

Ana.  G  =  100°  49'  30" 
C=  97°  26'  29" 
B  =  65°  33'  10" 

6.  Given  a  =  40°  5'  25"-6,  b  =  118°  22'7"'3,  ^  =  29°  42'33"'8; 
find  c,  (7  and  B. 

Ans.    c=153°38'42"-n  C   c=     90°    5' 41"-0 

C=160°    l'24"-4  I    or    J    C=    50°  18' 55"-2 
5  =    42°  37'  17"-5  J  [  -B  =  137°  22'  42"'5 

7.  Given  a  =  69°  34'  56",    b  =  120°  30'  30",    A  =  50°  10'  10"; 
find  c  and  C. 

^ln*.    c  — 70°  20'  20" 
C=50°30'    8"-4 

8.  Given  a  =  120°  30'  30",    6  =  69°  34'  56",     A  =  50°  10'  10" ; 
find  c  and  C. 

Ans.     Impossible. 

9.  Given  a  =  40°,  b  =  60°,  A  =  50° ;  solve  the  triangle. 

Ans.     Impossible. 

87.  CASE  III.  Given  a,  b  and  A.     Second  Solution.     We  find  B 
by  the  formula 

sin  A  sin  6 


sin  B  = 


sin  a 


and  then  by  Napier's  Analogies,  (41)  and  (43), 
cos 


, 


,    .      -, 

cot      C= 


or  by  (40)  and  (42), 


COS 


cos  ^  (  a  —  6  ) 


j. 

tan 


T>\ 

B} 


(154) 


sin  i  (A  -f-  5)  ,       .  /        ,  N 
tan  4  c  =  -  —f  tan  i  (a  —  6) 

sin  ^  (A  —  B) 


.^ 
cot  \  C 


,       1  /  A 
tan  i   ^1  — 


(155) 


sm  ^  (  a  —  6  ) 

in  which  we  employ  successively  the  two  values  of  B,  and  obtain 
two  solutions,  except  when  for  one  of  these  values  the  second  mem- 
bers become  negative,  for  ^  c  and  %  C  being  less  than  90°,  their 
tangents  must  be  positive. 


SOLUTION  OP  SPHERICAL  OBLIQUE  TRIANGLES.  197 

We  leave  it  to  the  student  to  apply  these  formulae  to  the  preceding 
examples. 

88.  To  determine  by  inspection  of  the  data  a,  6  and  A,  whether  there  are  two  solutions,  or 
but  one. 

1st.  It  has  already  been  seen,  Art.  85,  that  when  b  differs  more  from  90°  than  a, 
B  must  be  in  the  same  quadrant  as  6,  and  there  can  be  but  one  solution.  It  remains 
to  show, 

2d.  That  when  a  differs  more  from  90°  than  b,  there  will  necessarily  be  two  solu- 
tions. We  have,  by  the  first  of  (4), 

cos  a  —  cos  6  cos  c 

sin  c  = : — 

sin  6  cos  A 

Two  solutions  exist  so  long  as  both  values  of  c  are  positive,  and  less  than  180°,  that  is, 
so  long  as  sin  c  is  positive.  Now  when  a  differs  more  from  90°  than  6,  we  have,  (neg- 
lecting the  signs  for  a  moment), 

cos  a  >  cos  b  >  cos  b  cos  c 

herefore  the  numerator  of  the  above  value  of  sin  c  has  the  sign  of  cos  o.    But  by  Art. 

19,  VI.,  o  and  A  are  in  the  same  quadrant,  and  cos  a  and  cos  A  have  the  same  sign ; 
consequently  also,  the  numerator  and  denominator  have  the  same  sign,  and  the  value 
of  the  fraction,  or  of  sin  c,  is  positive,  as  was  to  be  proved.* 

Hence,  there  is  but  one  solution  when  the  side  opposite  the  given  angle  differs  less  from  90° 
than  the  other  given  side,  and  two  solutions  when  the  side  opposite  the  given  angle  differs  more 
from  90°  than  the  other  given  side. 

89.  CASE  IV.     Given  two  angles  and  a  side  opposite  one  of  them, 
or  A,  B  and  6.     (Fig.  9). 

First  Solution,  in  which  each  required 
part  is  deduced  directly  from  the  funda- 
mental formulae. 

To  find  c.     We  have,  by  (10), 

sin  c  cot  6  —  cos  c  cos  A  =  sin  A  cot  B 
or  multiplying  by  sin  6, 

sin  c  cos  b  —  cos  c  sin  6  cos  A  =  sin  A  cot  B  sin  b  (M) 

to  solve  which,  we  take 

k  sin  <p  =  sin  6  cos  A  ) 

V        (m) 

k  cos  <f>  =  cos  b  I 


*  The  same  proposition  may  be  otherwise  proved  thus.     By  the  equations  (m)  and 
(mf)  Art.  84,  we  have 


k  sin  0 

from  the  third  of  which  we  see  that  k  has  the  sign  of  cos  A  ;  if  then  a  differs  more 
from  90°  than  6,  that  is,  if  cos  a  and  cos  A  have  the  same  sign,  cos  <j>'  is  positive,  and 
$'  <  90°.  Also  since,  (neglecting  signs),  cos  a  >  cos  6,  we  have  cos  $f  >  cos  <f>,  or 
1>'  differs  more  from  90°  than  <f>.  Hence  0'  <  <t>  and  0'  <  180°  —  <f>,  or  ^  —  f  >  0 
and  0  -f-  0'  <  180°,  or  both  values  of  c  are  between  0  and  180°. 

R2 


(m') 


198  SPHERICAL  TRIGONOMETRY. 

then,  putting  c  —  (p  =  <pf,  (M)  becomes 

k  sin  <p'  —  sin  A  cot  B  sin  6 

or  eliminating  k,  we  have,  for  finding  c, 

tan  y  —  tan  6  cos  J. 

sin  ^>'  =  sin  tp  tan  ^4  cot  B 

C  =  (p  -\-  (D1 

Here  <pf  being  determined  by  its  sine,  will  have  two  values,  sup- 
plements of  each  other,  which  being  successively  added  to  y>,  give 
two  values  of  c. 

When  the  second  member  of  the  formula 

sin  <pr  =  sin  <p  tan  A  cot  B 

is  negative,  sin  <p',  and  therefore  <pr  is  negative,  and  the  two  supple- 
mental values  of  <p'  must  be  successively  subtracted  from  tp.     There 
will  be  two  solutions,  then,  except  when  one  of  the  values  of  c  exceeds 
180°,  or  when  one  of  them  is  negative. 
To  find  C.     We  have,  by  (5), 

sin  C  sin  A  cos  6  —  cos  C  cos  A  =  cos  B  (N) 

whence,  if  we  put 

h  sin  &  =  cos  A  ^ 

h  cos  &  =  sin  A  cos  6  j 

and  also  C  —  &  =  &',  we  have 

h  sin  &'  =  cos  5  ^ 

_  ,  (*') 

Eliminating  A,  we  have 

cot  $  =  tan  A  cos  6 

.     Q.      sin  &  cos  B 

sintf'  =  -  }.      (157) 

cosvl 


As  $'  is  also  determined  by  its  sine,  it  will  have  two  supplemental 
values,  which  will  both  be  added  to  or  both  subtracted  from  $,  (ac- 
cording to  the  sign  of  sin  $',)  thus  giving  two  values  of  C,  except 
when  one  of  them  exceeds  180°,  or  when  one  of  them  is  negative. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.            199 

To  find  a.     We  have  several  methods  :  1st,  directly  by  (3),  which 
gives 

sin  bsinA  n  ^ 

sin  a  =•  —  (15°) 
sin  B 

2d.  By  (146),  where  (p  and  <p'  have  the  same  values  as  in  this  case, 


COS  <£>'  COS  6 

cos  a  ~ 
3d.  By  (150), 


/-i  ~r>\ 
(Io9) 

cos  <p 


cot  a  =  cot  (pr  cos  B  (160) 

4th.  By  (147),  where  &  and  #'  have  the  same  values  as  in  this 
case, 

cos  &'  cot  6 
cot  a  =  — 

cos  # 

5th.  By  (152), 

cos  a  =  cot  #'  cot  5  (162) 

Each  of  the  last  four  formulae  gives  two  supplemental  values  of  a 
by  the  two  values  of  <p'  or  $',  employed  in  the  second  members. 
From  (156)  we  have 

cos  A  =  tan  <   cot  6 


which  with  (159)  gives 


cos  a  _  ,      sin  b 

-  —  cos  <p   X  ~i 
cos  A  sin  <p 


The  sign  of  the  second  member  of  this  equation  depends  upon  that 
of  cos  <p'  ',  since  sin  b  and  sin  <p  are  always  positive.  Hence  when 
cos  tp'  is  positive,  cos  a  and  cos  A  must  have  like  signs;  and  when 
cos  <f>'  is  negative,  cos  a  and  cos  A  must  have  different  signs.  A  like 
result  follows  from  the  first  of  (157)  and  (161)  with  reference  to  #'. 
Hence,  that  value  of  a  which  is  in  the  same  quadrant  with  A  belongs  to 
the  triangle  in  which  <pf  <C  90°,  $'  <  90°  ;  and  that  value  of  a  which 
is  in  a  different  quadrant  from  A  belongs  to  the  triangle  in  which 
<p'  >  90°,  &'  >  90°.  This  precept  enables  us  to  employ  (158)  with- 
out ambiguity.  In  the  use  of  (159),  (160),  (161),  and  (162),  it  is  only 
necessary  to  observe  the  algebraic  signs  of  the  several  terms. 

ChecJcs.  Of  the  various  formulae  above  given  for  finding  a,  one  or 
more  may  be  employed  for  the  purpose  of  verification.  When  c  and 
C  have  been  found,  however,  the  most  simple  check  is 


200  SPHERICAL  TRIGONOMETRY. 

sin  C sin  B 

sin  c        sin  6 


(163) 


which  might  have  been  employed  for  finding  C  after  c  was  found,  or 
reciprocally,  but  for  the  ambiguity  attaching  to  the  sines. 

90.  According  to  Art.  69,  VII.,  if  A  differs  more  from  90°  than 
B,  a  must  be  in  the  same  quadrant  with  A.      But  since  the  two 
values  of  a  are  supplements  of  each  other,  only  one  of  them  can 
satisfy  this  condition,  and  there  will  then  be  but  one  solution.     In 
such  case  c  and   C  will  each  be  found  to  have  but  one  admissible 
value. 

91.  The  problem  will  be  impossible  when  B  differs  more  from  90° 
than  A,  and  yet  is  not  in  the  same  quadrant  with  b.     In  such  case 
we  should  find  both  values  of  c  (and  both  values  of  (?)  to  be  greater 
than  180°,  or  both  negative. 

The  problem  will  also  be  impossible  when  sin  6  sin  A  >  sin  B, 
since  by  (158)  we  shall  then  have  sin  a  >  1. 


EXAMPLES. 

1.  Given  A  =  132°  16',  B  =  139°  44',  b  =  127°  30';  find  a. 

By  (158). 

B  =  139°  44'    0"  ar  co  log  sin  B  0-1895350 

A  =  132°  16'    0"  log  sin  A  9-8692449 

b  =  127°  30'    0"  log  sin    b  9*8994667 

a=    65°  16'  35"'l  log  sin  a  9-9582466 

or         a  =1U°  43'  24"-9 

2.  With  the  same  data,  find  C  and  a. 

By  (157). 

B=     139°  44'   0"  log  cos  £—9-8825499 

A=     132°  16'    O'Mogtanvl—  0-0414996  ar  co  log  cos  ^—0-1722547 
b  =     127°  30'    0"  log  cos  6—9-7844471 


56°  11'    8"-6  log  cot  #+  9-8259467          log  sin  #+9'91  95204 

70°  29'  31--OJ  sin*'+9^74327o 

109°30'29"-OJ 
126°  40'  39"-6 
165°  41'  37"-6 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  201 

By  (161).  Check.  (162). 

&  =   56°  11'    8"-6  ar  co  log  cos  $+0-2545328  log  cot  J3— 0-0720848 
&'=     70°29'31"-0) 
*V=  109°  30'  29"-0  j        logcos^'±9-5236676  log  cot  ^'±9j5  493427 

b  =127°  30'  0"  log  cot  b—  9-8849805  +  9-6214275 


log  cot  a +9-6631809  log  cos  a+ 9-6214275 
=     65°16'34"-9l 

Ans.  C=  126°  40'  39"-6  J          j  C=  165°  41'  37"-6 
a  =  114°  43'  25"'l  I          1   a  =    65°  16'  34"'9 

3.  Given  A  =  110°,  B  =  60°,  6  =  50°  ;  find  c  and  a. 

By  (156). 
£  =      60°  log  cot  £+9-9614394 

A=     110°  log cos^l— 9-5340517  logtan^l— 0-4389341 

b  =       50°  log  tan  b  +  0-0761865 

y=     157°49'26"-4     log  tan  <p—  9-6102382  log  sin  <p  +  9-5768627 

Cl=      121°    2'40"-2| 

c,=       14°36'12"-6J 

By  (159).  Check.  (160). 

<p  =     157° 49'26"-4arcologcos?>— 0-0333755  logcos£+9-6989700 

<p\=—  36°46'46"-2) 

/_      I^OOIO/IO//Q  r       logcos^'±9-90360301ogcot^'+0-1263669 
<p  2 — — 14o    lo  lo    "o  J 

b=     50°    0'    0"  log  cos  b  +9-8080675  +9-8253369 

f, 123°  46' 37"*5 1 

a=     56°13'22"-5J         log  cos  a +  9- 7450460  log  cot  a  +  9-8253369 

Jns.      c  =  121°    2'40"-2|  f  c  =  14°  36'  12"-6 

a  =  123°46'37"-5J          {  a  =  56°  13'  22"-5 

4.  Given  A  =  60°,  B  =  110°,  6  =  50° ;  find  c. 

Ans.    o  =  11°  57'  47"-9 

5.  Given  A  =  115°  36'  45",  £  =  80°  19'  12",  b  =  84°  21'  56"; 
find  a,  c  and  C. 

^w.s.  a  =114°  26' 50" 
c—  82°  33' 31" 
C=  79°  10' 30" 

6.  Given  A  =  61°  37'  52"-7,  £=139°  54'  34"-4,  b  =  1 50°  17'  26"-2 ; 
find  o,  c  and  C. 

Ans.     o=    42°37'17"-5-)  r   o=  137°  22'  42"-5 

c  =  129041'    4"-8  V    or    <?     c-     19°58'35"-6 

C==    89°54'19"-OJ  I  C=    26°21'17"-6 

26 


202  SPHERICAL  TRIGONOMETRY. 

7.  Given  A  =  70°,  B  =  120°,  b  =  80°  ;  solve  the  triangle. 

Ans.     Impossible. 

8.  Given  A  =  60°,  B  =  40°,  6  =  50° ;  solve  the  triangle. 

Ans.     Impossible. 

92.  CASE  IV.  Given  A,  B  and  6.     Second  Solution.     We  find  a 
by  the  formula 

sin  b  sin  A 


sin  a  = 


sin  B 


and  then  by  Napier's  Analogies  we  find  c  and  C,  precisely  as  in 
Case  III.,  Art.  87,  employing  successively,  in  (154)  or  (155),  the  two 
values  of  a  given  by  the  preceding  equation.  There  will  be  but  one 
solution,  if  one  of  these  values  renders  the  second  members  of  (154) 
or  (155)  negative. 

The  student  should  apply  this  method  to  the  preceding  examples. 

93.  To  determine  by  inspection  of  the  data  A,  B  and  b,  whether  there  are  two  solutions  or 
but  one. 

1st.  It  lias  already  been  seen,  Art.  90,  that  when  A  differs  more  from  90°  than  B, 
a  must  be  in  the  same  quadrant  with  A,  and  there  can  be  but  one  solution.  It  remains 
to  show  that, 

2d.  When  B  differs  more  from  90°  than  A}  there  will  necessarily  be  two  solutions. 
We  have,  by  (5), 

.     ^, cos  B  -\-  cos  A  cos  C 

sin  o  —  — -     ;     -         - 
sin  A  cos  o 

Two  solutions  exist  so  long  as  both  values  of  C  are  less  than  180°,  and  both  positive, 
that  is,  so  long  as  sin  C  is  positive.  Now  wlien  B  differs  more  from  90°  than  A,  we 
have,  (neglecting  signs  for  a  moment)," 

cos  B  >  cos  A  >  cos  A  cos  C 

therefore  the  -numerator  of  the  value  of  sin  C  has  the  sign  of  cos  2?.  But  by 
Art.  69,  VII.,  B  and  6  are  in  the  same  quadrant,  consequently  the  numerator  and 
denominator  have  the  same  sign,  and  the  value  of  the  fraction,  or  of  sin  C  is  always 
positive,  as  was  to  be  proved.* 

Hence,  there  rs  but  one  solution  when  the  angle  opposite  the  given  side  differs  less  from 
90°  than  the  other  given  angle ;  and  two  solutions  when  the  angle  opposite  the  given  side 
differs  more  from  90°  than  the  other  given  angle. 

94.  CASE  IV.  might  have  been  reduced  to  Case  III.  by  means  of  the  polar  triangle 
of  Art.  8.      For  there  will   be  known  in  the  polar  triangle  two  sides  and  an  angle 
opposite  one  of  them,  being  the  supplements  of  the  given  angles  and  side  of  the  pro- 
posed triangle.     The  polar  triangfe  being  solved,  therefore,  by  Case  III.,  and  its  two 
remaining  angles  and  third  side  found,  the  supplements  of  these  parts  will  be  the 
required  sides  and  third  angle  of  the  proposed  triangle. 


*  It  may  be  shown  that  both  values  of  C  will  be  admissible,  by  a  process  of  reason- 
ing similar  to  that  employed  in  the  note  on  page  197,  applied  to  the  equations 
of  Art.  89. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES. 

95.  CASE  V.   Given  the  three  sides,  or  a,  b  FlG- 9- 

and  c.    (Fig.  9.)     We   have  three  methods 
for  computing  the  half  angles  : 

1st.  By  the  sines,  from  (31),  remembering 
that 

s  =  ^  (a  +  6  -f-  c) 

.     ,  /  /  sin  (s  —  6)  sin  (s  —  c)  \ 

sm  %  A  =  \  I  [ 

*  \  sin  6  sin  c  / 

sin  £  B  =  J  (  S1"  '  T  C  — S  } 

*  \  sin  c  sin  a  / 


203 


sin 


A  \  sin 

i  n  —     I  (  sm  (s  —  a)  sm  (s  —  b) 


C 


_      /  / 


** '  sin  a  sin  6  / 


2d.  By  the  cosines,  from  (33), 


cos     - 


cos 


cos 


//  sin  s  sin  (s  —  a)  \ 
sin  6  sin  c       / 


/  /       *  "        /  J\\   \ 

'  \       sin  c  sin  a       / 

iC  =  J/2El_!SJ«_! 

^  \       sin  a  sin  6 


3d.  By  the  tangents,  from  (34), 


|  /  sin  (g  —  6)  sin  (s  —  c) 
Al  \       sin  s  sin  (s  —  a) 

/  /  sin  (s  —  c)  sin  (s  —  a)\ 
tan  4  jS  =  A  / 1 

.     '  \       sm  s  sin  (s  —  6)      / 

\  i  ,'    /          \   '    (        j\\  \ 

tan  ^  C  =  \  i  -  /  , — * — r—1  I 

V  \      sm  s  sin  (s  —  c)       / 


(164) 


(165) 


(166) 


When  only  one  of  the  angles  is  required,  the  simplest  method  will 
be  by  (16,5),  but  if  the  required  uugle  is  less  than  90°,  it  will  be 
found  more  accurately  by  (164),  for  then  ^  A  <  45°,  and  the  sine 
varies  more  rapidly  than  the  cosine.  And,  for  a  similar  reason,  if 
the  angle  is  greater  than  90°,  we  should  prefer  (165).  By  (166)  we 
always  have  an  accurate  result,  although  the  formula  is  not  quite  so 
simple. 

When  the  three  angles  are  required,  (166)  will  require  the  least 
labor,  since  sin  a,  sin  6,  and  sin  c,  are  not  then  required. 


204  SPHERICAL  TRIGONOMETRY. 

No  ambiguity  can  arise  in  these  solutions,  since  the  half  angles 
must  be  less  than  90°  ;  they  require  therefore  no  attention  to  the 
algebraic  signs. 

EXAMPLES. 

1.  Given  a  =  100°,   6  =  50°,   c  =  60°  ;  find  A. 

a  =100° 

b=    50°  logcosec  0-1157460 

c=    60°  logcosec  0-0624694 
2s  =  210° 

s=105°  log  sin  9.9849438 

8  —  a  =      5°  log  sin  8*9402960 

2)9-1034552 

%A=   69°    7' 52"-7  log  cos  9-5517276 
A  =  138°  15'  45"-4 

2.  With  the  same  data,  find  all  the  angles. 

By  (166). 

a  =  105°  1.  cosec  0-0150562  1.  cosec  0-0150562  1.  cosec  0-0150562 

s  —  a=      5°  1.  cosec  1-0597040      1.  sin  8-9402960     1.  sin  8-9402960 

*  —  b  =    55°      1.  sin  9-9133645  1.  cosec  0-0866355     1.  sin  9-9133645 

8  _  c  =    45°      1.  sin  9-8494850      1.  sin  9-8494850  1.  cosec  Q-l  505150 

2)0-^376097  2)8-8914727          2)9-01 923T7 

1.  tan  0-4188049     1.  tan  9-4457364     1.  tan  9-5096159 

Ans.  ^  =  138°  15'  45"-4     B  =  31°  11'  14"-0     C  =  35°  49'  58"-2 

3.  Given  a  =  10°,   6  =  7°,    c  =  4°  ;  find  the  angles. 

Ans.  ^4=128°  44'  45"- 1 
B=  33°  11'  12"-0 
C=  18°  15'31"-1 

96.  The  method  by  (166)  may  be  put  under  the  following  convenient  form.     Let 

p I  /sin  (s  —  a)  sin  (a  —  b)  sin  (g  —  c)\ 

\  V  sin  *  / 

then  (167) 

UnM  =  -— ^ rt     tan*£  =  -  — ,     tanJC  = 

sin  (s  —  a)  sin  (s  —  6)  sin  [»  —  c) 

which  are  similar  to  the  formulae  of  PI.  Trig.  Art.  146,  and  are  computed  in  the  same 
manner. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES. 


205 


97.  CASE  V.  Given  a,  b  and  c.     Second  Solution.    If  the  whole  angle  is  required 
directly,*  we  have 

A  _  cos  a  —  cos  b  cos  c 
sin  6  sin  c 

which  may  be  adapted  for  logarithms  by  an  auxiliary  thus  : 
cos  </>  =  cos  b  cos  c 


Or  thus, 


sin  6  sin  c 


.   .       cos  b  cos  c 

COt  9  =  ; — 

sin  a 


(169) 


FIG.  9. 


sin  b  sin  c  sin  <t> 

98.  CASE  VI.   Given  the  three  angles,  or 
A,  B  and  C.  (Fig.  9).  We  have  three  methods 
of  finding  the  half  sides : 

1st.  By  the  sines,  (36). 
2d.  By  the  cosines,  (38). 
3d.  By  the  tangents,  (39). 
The  computations  are  conducted  precisely  in  the  same  form  as  those 
of  the  preceding  case. 

EXAMPLE. 

Given  A  =  120°,  B  =  130°,  C=  80° ;  find  c. 

Ans.  c  =  41°  44'  14"-6 

99.  The  formulae  (39)  may  be  arranged  for  convenient  use  in  the  same  manner  as 
the  corresponding  formulae  of  the  preceding  case,  Art.  96. 

100.  CASE  VI.  Given  A,  B  and  C.    Second  Solution.     We  have,  by  (5), 

cos  A  -4-  cos  B  cos  C 

cos  a  = 7-1 — 

sin  B  sin  C 

which  may  be  adapted  for  logarithms  by  an  auxiliary,  thus: 
cos  <t>  —  cos  B  cos  0 

(170) 


or, 


tan  0  — 


sin  B  sin  C 
cos  B  cos  C 


sin  A 

cos  (A  — 


sin  B  sin  O  cos  0 


(171) 


*See  NOTE  at  the  end  of  this  chapter,  p.  211,  for  the  method  of  computing  many 
of  the  general  formulie  of  spherical  trigonometry  directly,  without  the  aid  of  auxiliary 
angles. 

S 


206  SPHERICAL  TRIGONOMETRY. 


101.  All  the  cases  of  oblique  spherical  triangles  may  be  solved  by  dividing  the 
triangle  into  two  right  triangles  by  a  perpendicular  from  one  of  the  vertices  to  the 
opposite  side,  and  solving  these  partial  triangles  by  the  methods  of  the  preceding 
chapter.  Bowditch  has  given  two  rules,  based  upon  Napier's  Rules,  (Art.  46),  by 
which  the  application  of  this  method  is  facilitated. 

FlG-  10-  102.  Bowditch' s  Rules  for  Oblique  Triangles.     "  If  in  a 

spherical  triangle,  (Fig.  10),  two  right  triangles  are 
formed  by  a  perpendicular  let  fall  from  one  of  its  ver- 
tices upon  the  opposite  side ;  and  if,  in  the  two  right 
triangles,  the  middle  parts  are  so  taken  that  the  perpen- 
dicular is  an  adjacent  part  in  both  of  them ;  then 

The  sines  of  the  middle  parts  in  the  two  triangles  are 
proportional  to  the  tangents  of  the  adjacent  parts. 
But  if  the  perpendicular  is  an  opposite  part  in  both  the  triangles,  then 
The  sines  of  the  middle  parts  are  proportional  to  the  cosines  of  the  opposite  parts. 
To  prove  which  rules,  let  M  denote  the  middle  part  in  one  of  the  right  triangles, 
A  an  adjacent  part,  and  0  an  opposite  part.     Also,  let  m  denote  the  middle  part  in 
the  other  triangle,  a  an  adjacent  part,  and  o  an  opposite  part;  and  let  p  denote  the 
perpendicular. 

First.  If  the  perpendicular  is  an  adjacent  part  in  both  triangles,  we  have,  by  Napier's 
Rules,  (Art  46,) 

sin  M  =  tan  A  tan  p 
sin  m  =  tan  o  tan  p 
whence 

sin  M tan  A  tan  p  _  tan  A 

sin  m       tan  o  tan  p      tan  a 

or  sin  M:  sin  m  =  tan  A  :  tan  a 

Secondly.  If  the  perpendicular  is  an  opposite  part  in  both  triangles,  we  have,  by 
Napier's  Rules 

sin  M  —  cos  O  cos  p 
sin  m  =  cos  o  cos  p 
whence 

sin  M cos  0  cos  p cos  O 

sin  m      cos  o  cos  p        cos  o 

or  sin  M :  sin  m  =  cos  0  :  cos  o"  * 

We  proceed  to  solve  the  six  cases  of  spherical  triangles  with  the  aid  of  a  perpen- 
dicular. It  will  be  seen,  however,  that  Bowditch's  Rules  are  applicable  but  in  the  first 
four  cases. 

*Peirce's  Spherical  Trigonometry,  Art.  44. 


SOLUTION   OF  SPHERICAL  OBLIQUE  TRIANGLES.  207 

103.  CASE  I.  Given  b,  c  and  A.  Let  the  perpendicular  C  P,  Fig.  10,  be  drawn 
frcm  C,  (that  is,  in  such  a  manner  as  to  put  two  given  parts  in  one  of  the  right  tri- 
angles). Then  the  right  triangle  A  C  P  gives,  by  Napier's  Rules,  if  we  put  A  P  =  0, 

tan  (j>  =  tan  6  cos  A  (172) 

then   taking   co.  6   and    co.  a  as  middle   parts   in   the  two  triangles,  A  P=<t>  and 
B  P  =  c  —  0  *  are  the  opposite  parts,  whence,  by  Bowditch's  Rules, 

cos  <t>  :  cos  (c  —  <j>)  =  cos  6  :  cos  a 
whence 


cos? 

Again,  taking  A  P  and  P  B  as  middle  parts,  co.  A  and  co.  B  are  adjacent  parts, 
whence,  by  Bowditch's  Rules, 

sin  0  :  sin  (c  —  <f)  =  cot  A  :  cot  B 
whence 


and  the  formulae  (172),  (173),  (174),  agree  entirely  with  (122)  and  (123). 
The  triangle  B  C  P  gives  as  a  check 

tan  a  cos  B  =  tan  (c  —  tf>)  (  175) 

which  agrees  with  (124). 

By  drawing  the  perpendicular  from  B,  we  may  in  the  same  manner  obtain  the 
formulae  (125). 

The  angle  C  may  be  found  by  the  proportion 

sin  a  :  sin  c  =  sin  A  :  sin  C 

or  if  C  has  been  found  by  means  of  a  perpendicular  from  B,  B  may  be  found  by  a 
similar  proportion,  as  in  Art.  72  ;  and  the  quadrant  in  which  the  angle  is  to  be  taken 
must  be  determined  by  the  principles  of  Art.  69. 

104.  CASE  II.    Given  A,  G  and  b.      Let  the  perpendicular  be  drawn  as  before, 
Fig.  10,  and  let 


then,  by  Napier's  Rules, 

cot  &  =  tan  A  cos  6  (176) 

and  by  Bowditch's  Rules,  taking  co.  A  and  co.  B  as  middle  parts,  and  therefore 
co.  A  C  P  and  co.  B  C  P  as  opposite  parts, 

sin  1?  :  sin  (C  —  #)  =  cos  A  :  cos  B 
whence 

sin  (C  —  #)  cos  A  /i77\ 

=  - 


*If  A  P  should  exceed  A  B,  (that  is,  if  the  perpendicular  should  fall  without  the 
triangle),  BP  would  be  equal  to  AP  —  A  B  =  <j>  —  c,  and  the  solution  could  be 
modified  accordingly.  But  the  true  results  will  always  be  obtained  by  regarding  B  P 
as  negative  ;  that  is,  by  still  taking  B  P=c  —  <t>  and  attending  to  the  signs  of  all  the 
terms  as  already  exemplified,  p.  182. 

flf  A  C  P>  A  C  B,  B  CP=  C—A  C  P  will  become  negative,  but  the  true 
results  are  still  found  by  attending  to  the  signs,  as  already  shown,  p.  187. 


208 


SPHEKICAL  TRIGONOMETRY. 


Again,  taking  co.  A  CP  and  co.  -B  CP  as  middle  parts,  and  therefore  co.  b  and  co.  o 
as  adjacent  parts,  Bowditch's  Rules  give 

cos  & :  cos  (C —  #)  =  cot  b :  cot  a 
whence 

/  X^  O\  «     1 

(178) 


and  (176),  (177),  (178),  agree  entirely  with  (134)  and  (135). 
The  triangle  B  C  P  gives 

tan  ^  cos  a  =  cot  (C—  #)  (179) 

which  agrees  with  (136). 

By  drawing  the  perpendicular  from  A,  we  may  in  the  same  manner  obtain  the 
formula  (137). 

The  side  c  may  be  found  from  the  proportion 

sin  A  :  sin  C=  sin  a  :  sin  c 

and  Art.  69;  or  c  being  found  by  means  of  a  perpendicular  from  A,  we  may  find  o  by 
a  similar  proportion. 

105.  CASE  III.  Given  a,  b  and  A.  Let  the  per- 
pendicular be  drawn  from  C,  Fig.  10,  as  in  the  preced- 
ing cases,  and  let  A  P  =  <j>,  £  P  =  <$>' ;  then,  by  Na- 
pier's Kules, 

tan  0  =  tan  b  cos  A  (180) 

and,  by  Bowditch's  Rules, 

cos  b  :  cos  a  =  cos  ?  :  cos  <j>' 


FIG.  10. 


whence 


and  then 


cos  r  = 


cos  6 


(181) 


In  Art.  84,  we  have  found,  from  analytical  considerations,  that  this  case  admits  of 
two  solutions,  and  that  the  general  expression  for  c  is 

c  =  0=h^/  (182) 

In  fact,  let  us  attempt  to  construct  the  triangle  with  the  data  a,  b  and  A.    Having 
constructed  A  equal  to  the  given  angle,  and  b  equal  to  the  adjacent  side,  Fig.  11,  let 
•  !!•  a  small  circle  be  described  about  C  as  a  pole, 

with  a  (circular)  radius  =  a;  this  circle  in- 
tersects the  great  circle  A  B  in  two    points, 
B  and   B',   and   both    triangles,  A  CB  and 
A  CB'  contain  the   same  data  a,  6  and  A. 
If  the  perpendicular  CP  is  drawn,  we  have 
BP  —  B'P,  so  that  in  one  of  the  triangles, 
the  side  c  =  AB  =  A  P  +  P  B  =  <f>  +  0', 
and  in  the  other,  c  =  A  B'  =  A  P  —  B'  P  =  0  —  f .     If  both  points  of  intersection, 
B  and  B',  fall  on  the  same  side  of  A,  and  within  180°  of  A,  both  solutions  will  be 
admissible. 
To  find  C,  let  A  CP=  #,  and  BCP=  #',  then  by  Napier's  Rules, 


cot  &  =  tan  A  cos  b 


(183) 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  209 

and  by  Bowditch's  Kules, 

cot  6  :  cot  a  =  cos  i?  :  cos  $' 
whence 

cos  $'  —  cos  &  tan  b  cot  a  (184) 

and  since  in  Fig.  11, 
C=ACB  =  ACP+BCP=d  +  #',  or  C=ACB'  =  ACP—  B'CP=$  —  tf', 

we  have 

C=$±$'  (185) 

and  the  formula:  (180),  (181),  (182),  (183),  (184),  (185),  agree  entirely  with  (146)  and 
(147). 
After  c  was  found,  we  might  have  found  C  from  the  proportion 

sin  a  :  sin  c  =  sin  A  :  sin  C  (186) 

and  B  is  found  from  the  proportion 

sin  a  :  sin  6  =  sin  A  :  sin  B  (187) 

The  two  values  of  B  determined  by  (187),  are  both  admissible  when  c  has  two  values 
as  above.  It  is  also  evident,  from  Fig.  11,  that  the  two  values  of  B  are  supplemental. 
To  determine  the  corresponding  values  of  c  and  B,  we  observe  that,  by  Art.  49,  the 
perpendicular  OP  is  in  the  same  quadrant  with  A  and  -with  OBP  and  CB'P,  and 
therefore  CB'A  is  in  a  different  quadrant  from  A.  Hence,  that  value  of  B  which  is  in 
the  same  quadrant  as  A  corresponds  to  the  value  of  c  =  <j>  +  ^,  and  that  value  of  B  which  is 
in  a  different  quadrant  from  A  corresponds  to  the  value  of  c  =  <t>  —  <j>'  ;  which  agrees  with 
what  is  shown  in  Art.  84. 

In  computing  (186),  the  two  values  of  c  must  be  employed  successively,  and  the 
formula  computed  twice.  At  each  computation  we  shall  have  two  values  of  C  found 
from  the  sine,  one  of  which  must  be  selected  by  Art.  69.  But  as  the  application  of  the 
principles  of  Art.  69  is  tedious  and  embarrassing,  it  is  better  to  find  C  by  (184) 
and  (185). 

The  formulae  (149),  (150),  (151),  (152),  for  finding  B,  may  easily  be  deduced  by 
Napier's  and  Bowditch's  Rules. 

106.  CASE  IV.  Given  A,  B  and  b.  Let  the  perpendicular  be  drawn  as  before, 
Fig.  10,  and  let  AP  =  <t>,  BP=<p',  then  as  before, 

tan  ^  =  tan  6  cos  A  (188) 

and  by  Bowditch's  Rules, 

cot  A  :  cot  B  =  sin  0  :  sin  <t>' 
whence 

sin  </  =  sin  <j>  tan  A  cot  B 


which  agree  with  (156).     But  <]/  having  two  supplemental  values  determined  by  the 
sine,  c  has  two  values,  as  already  explained  in  Art. 

To  show  the  same  geometrically,  let  BP, 
Fig.  12,  be  the  acute  value  of  $'  ',  and  about 
C  as  a  pole,  let  a  small  circle  be  described 
passing  through  B,  and  intersecting  the 
great  circle  AB  again  in  B".  Let  BffC  be 
drawn,  and  produced  to  meet  AB  again  in 
Bf,  forming  the  lime  B"B'.  Then  we  have 

B'  =  B"  =  CBA 

27  s2 


210  SPHERICAL  TRIGONOMETRY. 

so  that  in  both  triangles,  ACB  and  ACS', 
the  value  of  the  angle  opposite  the  side  b  is 
the  same,  that  is,  both  triangles  contain  the 
same  data,  A,  JB  and  6.  Now 

180°  =  B"B'=  B';P  +  B'P=  BP  +  B'P, 
so  that  .BPand  B'P  are  supplements  of  each 
other. 
In  the  triangle  A  CB  we  have 


and  in  the  triangle  A  CB'  we  have 

=  AP+B'P 


and  hence  the  two  values  of  c  are  found  by  giving  $'  its  acute  and  obtuse  values  suc- 
cessively, as  already  shown  analytically. 

By  Art.  49,  CP  must  be  in  the  same  quadrant  with  A;  hence,  if  B  is  in  the  same 
quadrant  with  A,  P  falls  between  A  and  B,  as  in  the  figure,  and  for  the  same  reason, 
between  A  and  B'.  But  if  A  and  B  were  in  different  quadrants,  both  points,  B  and  B', 
might  fall  between  A  and  P.  The  two  values  of  c  would  then  be  found  by  the 
formula 

c  =  0-/ 

0'  taking,  successively,  its  acute  and  obtuse  values.     In  that  case,  tan  A  and  cot  B 
would  have  opposite  signs  in  (189),  sin  <j>'  would  be  negative,  which  would  make  0' 
negative,  so  that  the  true  results  will  be  obtained,  without  reference  to  a  diagram,  by 
attending  to  the  signs  of  the  several  terms,  as  already  fully  exemplified,  p.  201. 
To  find  C;  let  ACP=  &,  BCP=$',  then  we  have,  as'  before, 

cot  $  =  tan  A  cos  6  (190) 

and  by  Bowditch's  Rules 

cos  A  :  cos  B  =  sin  •&  :  sin  &' 
whence 


.     „.      sin  #  cos  B 
sm  $'  =  — 

cos  A 


(191) 


which  agree  with  (157).     It  is  evident  from  Fig.  12,  that  BCP  and  B'CP,  are  sup- 
plemental, and  that  the  remarks  above  made  with  reference  to  <f>'  apply  also  to  &'. 
After  c  was  found,  we  might  have  found  C  by  the  proportion 

sin  6  :  sin  c  =  sin  B  :  sin  C  (192) 

and  a  is  found  by  the  proportion 

sin  B  :  sin  A  =  sin  6  :  sin  a  (193) 

The  two  values  of  a  found  by  (193)  are  both  admissible  when  c  has  two  values. 
From  Art.  50,  it  follows  that  when  BP  is  acute,  a  must  be  in  the  same  quadrant 
with  CP,  that  is,  (Art.  49),  in  the  same  quadrant  with  A ;  and  when  BP  is  obtuse, 
o  must  be  in  a  different  quadrant  from  A.  That  is,  that  value  of  a  vihich  is  in  thi'  .-unm: 
quadrant  with  A,  belongs  to  the  triangle  in  which  0'  <  90°,  and  that  value  of  a  which  is  in  a 
different  quadrant  from  A,  belongs  to  the  triangle  in  which  0'  >  90°  ;  which  agrees  with 
Art.  89. 

The  formulae  (159),  (160),  (161),  and  (162),  for  finding  a  may  easily  be  deduced  by 
Napier's  and  Bowditch's  Rules. 


SOLUTION  OF  SPHERICAL  OBLIQUE  TRIANGLES.  211 

107.  CASE  V.  Given  a,  b  and  c.  The  perpendicular 
cannot  be  drawn,  in  this  case,  so  that  two  of  the  given 
parts  shall  be  in  one  triangle  ;  nevertheless  the  case  can 
be  solved  by  means  of  a  perpendicular.  Let  the  perp. 
be  drawn  from  any  angle,  as  C,  Fig.  13,  and  as  before, 
put  AP=<f>,  EP  =  <!>';  then  by  Bowditch's  Rules, 

cos  0  :  cos  $'  =  cos  6  :  cos  o 

whence  cos  $'  —  cos  ft  _  cos  a  —  •  cos  6 

cos  0'  -f-  cos  ft      cos  a  +  cos  6 
or,  by  PI.  Trig.  (110), 

tan  i  (<j>  +  f)  tan  $  (<t>  —  <j>')  =  tan  $  (b  -f  a)  tan  J  (6  —  a) 
whence,  since  <j>  -f-  $'  =~-  c, 

tan  J  (ft  —  ft')  =  tan  $  (b  +  a)  tan  £  (6  —  a)  cot  J  c  1     /  1  Q^\ 

=    «  /    ' 


which  determine  £  (ft  —  ft')  and  £  (ft  +  ft')  whence  ft  and  ft'.     The  angles  A  and  £ 
are  then  determined  by  Napier's  Rules. 

108.  CASE  VI.   Given  A,  B  and  C.     In  Fig.  13,  let  ACP=&,  ECP=&';  then, 
by  Bowditch's  Rules, 

sin  $  :  sin  #'  =  cos  J.  :  cos  B 
whence 

sin  1?  —  sin  #'  _  cos  A  —  cos  B 
sin  #  -)-  sin  i9/       cos  A  -f-  cos  5 

or,  by  PI.  Trig.  (109)  and  (110), 


whence,  since  #  +  i?r  =  (7, 

tan  i  (#  —  *0  =  tan  ±(B  +  A)  tan  J  (B  —  A)  tan  |  C7 

}(*  +  ^)  =  iC7 

which  determine  $  (#  —  ^r)  and  i  (i^  +  ^x)  an<l  therefore  i^  and  <9/.    The  sides  a  and  6 
are  then  found  by  Napier's  Rules. 

NOTE   REFERRED  TO   ON   PAGE   205. 
Computation  of  Spherical  Formulae  by  the  Gaussian  Table. 

The  Gaussian  Table  is  a  table,  first  suggested  by  Gauss,  for  readily  computing  the 
logarithm  of  the  sum  or  difference  of  two  quantities,  when  the  logarithms  of  these 
quantities  are  given. 

If  p  and  q  are  the  two  numbers  whose  logarithms  are  given,  p  being  the  greater 
number,  (or  log  p  the  greater  logarithm),  we  have,  in  the  first  place 


If,  then,  we  put  x  =  %-,  we  have 
9 

log  x  =  log  p  —  log  q 

log  (  p  +  q)  =  log  q  +  log  (1  -f  *) 
or  log  ( p  +  q)  =  log  p  +  log  (l  +  —  \ 


212  SPHERICAL  TRIGONOMETRY. 

Downes's  Table  XXII.,  with  the  argument  log  x,  the  difference  of  the  given  logar- 
ithms, gives  log  (1  +  x),  which  being  added  to  log  q,  the  less  logarithm,  gives  the 
required  log.  sum,  or  log(p-\-q).  Table  XXIII.,  with  the  argument  log  x,  gives 

logll  + |  which,  being  added  to  log  p,  the  greater  logarithm,  gives  the  required 

log.  sum.     Either  table  may,  in  general,  be  employed,  but  one  or  the  other  may  be 
found  more  convenient  in  a  particular  application,  and  therefore  both  are  given. 
Again,  we  have 


so  that,  putting,  as  before,  x  =  1L,  we  have 

9 

log  x  —  log  p  —  log  q 
log  (  P  —  q)  =  log  p  +  log  (l  —      j 

Downes's  Table  XXIV.,  with  the  argument,  log  x,  gives  log  II  —  _|  which,  being 

added  to  the  greater  logarithm,  gives  the  required  log.  difference,  or  log  ( p  —  q), 

With  these  tables,  then,  we  may  readily  compute  any  of  the   preceding  formulae 
which  contain  two  terms  in  the  second  member,  without  the  aid  of  auxiliary  angles. 

EXAMPLES. 

1.  Given  6  =  120°  3<X  30",    c  =  70°  20'  20",    A  =  50°  10'  10";  find  a.    (Same 
as  Ex.  1.  p.  182). 
The  formula  is 

cos  a  =  cos  6  cos  c  -f-  sin  6  sin  c  cos  A 

which  will  be  thus  computed : 

log  cos  b  —  9.70557 
log  cos  c  +  9-52693 

log  q  —  9-23250 

log  sin  6  +  9-93529 
log  sin  c  +  9-97391 
log  cos  A  +  9-80654 

log  p  +  9-71574 
log  p  —  log  q  =  log  x  —  0-48324 

The  terms  p  and  q  have  opposite  signs,  and  although,  by  the  formula,  they  are  to  be 
added  (algebraically),  an  arithmetical  difference  is  required.  By  marking  the  signs 
of  all  the  quantities,  as  above,  we  shall  always  know  whether  a  sum  or  difference  is 
required  by  the  sign  before  log  x.  In  this  case  this  sign  being  negative,  we  are  to  find 
a  difference,  and  therefore,  by  Table  XXIV.,  we  take 

log  (l  -  -)        9-82694 

log  p  +  9-71574 

log  cos  a  +  9-54268 

a  =  69°  34'  52" 


SOLUTION    OF    SPHERICAL    OBLIQUE    TRIANGLES.          213 

2.  With  the  same  data,  find  B.     The  formula  is 

.   T, sin  c  cot  6  —  cos  c  coa  A 

sin  A 

which  must  here  be  put  under  the  form 


sin  A 

and  is  thus  computed  : 

log  sin  c  +  9-97391 

log  cot  b  —  9-77029 

ar  co  log  sin  A  -j-  0*11467 

log  p  —  9-85887 

log  (—cos  c)  —  9-52693 
log  cot  A  +  9-92121 

log  q  —  9-44814 
log  p  —  log  q  =  log  x  +  0-41073 

where  the  sign  before  log  x  being  positive,  the  tables  for  log.  sum  must  be  used.    By 
Table  XXII.,  we  have 

log  (1  +  x)       0-55325 

(which  is  to  be  added  to  the  less  log.)  log  q  — 9*44814 

log  cot  B  —  0-00139 
or,  by  Table  XXIII. ,  we  have 

\--\       0.14252 


(which  is  to  be  added  to  the  greater  log.)  log  p  —  9'85887 

log  cot  B  —  0-00139 
B  =  135°  5'  31" 

In  these  isolated  examples,  the  labor  of  computation  is  very  little  less  than  with 
the  use  of  an  auxiliary  angle,  as  on  p.  182;  but  the  Gaussian  Table  has  greatly  the 
advantage  when  the  same  formula  is  to  be  repeatedly  computed  with  successive 
values  of  one  of  the  data  while  the  others  remain  constant.  Thus,  in  the  first  of 
the  preceding  examples,  if  successive  values  of  a  are  to  be  found  corresponding  to 
successive  values  of  A,  while  b  and  c  are  constant,  log  q  will  be  constant,  and  log  x 
will  take  successive  values,  corresponding  to  those  of  log  cos  A,  so  that  after  the 
first  value  of  a  is  found  the  succeeding  ones  are  rapidly  obtained.  On  the  other  hand, 
as  the  auxiliary  0  in  the  formulae  (122),  depends  upon  A,  the  whole  process  would 
have  to  be  repeated  in  finding  each  value  of  a. 

For  other  forms  of  the  Gaussian  Table,  see  the  original  table,  (to  five  places  of 
decimals),  by  Gauss,  published  in  Zach's  Monatliche  Coi-respondenz,  Nov.  1812 ;  Mat- 
thiessen's,  (to  seven  places),  Altona,  1817  ;  in  Vega's  Sammlung  mathematischer  Ta- 
fdn,  (five  places),  Leipzig,  1840;  Zech,  (seven  places),  Leipzig,  1849;  Shortrede's 
Collection  of  Tables,  (seven  places),  Edinburgh,  1849  ;  Gray's  Tables  for  the  Compu- 
tation of  Life  Contingencies,  (six  places),  London,  1849  ;  Schumacher's  Hiilfstafeln, 
new  ed.  (four  places). 


CHAPTER   IV. 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE. 

109.  WE    have   thus   far,  following  the   usual  course,  considered  those  spherical 
triangles  only  whose  sides  and  angles  are  less  than  180°.     In  the  applications  of  this 
subject   in    astronomy,   however,  it   is    often  necessary  to  consider  triangles  whose 
sides  or  angles  exceed  180°.     (For  example,  the  right  ascension  of  a  heavenly  body, 
admitting  of  all  values  from  0°  to  360°,  may  be  one  part  of  such  a  triangle).     We 
may,  it  is  true,  in  such  cases,  always  substitute  another  triangle  whose  parts  are  the 
supplements  to  180°  or  360°   of  those  of   the  proposed   triangle  ;  but   this   mode, 
although  very  generally  regarded  as  the  simplest,  is  not  really  so  in  the  cases  alluded 
to.     The  construction  of  figures  for  discovering  the  supplemental  triangles  is  often 
embarrassing  and  liable  to  mistake,  while  the  solutions,  when  obtained,  are  mostly 
deficient  in  generality,  and  can  only  be  regarded  as  solutions  of  the  particular  cases 
of  a  general  problem.     But  if  we  proceed  by  a  method  that  is  as  applicable  when  the 
parts  of  the  triangle  exceed  as  when  they  are  less  than  180°,  we  may  investigate  a 
problem  under  the  simplest  supposition  of  the  values  of  these  parts,  and  rely  upon  the 
generality  of  the  method  to  cover  all  the  particular  cases. 

110.  We  shall  first  endeavor,  in  an  elementary  manner,  to  give  the  student  a  con- 
ception of  the  nature  of  the  general  spherical  triangle. 

FIG.  14. 

Let  ABC,  Fig.  14,  be  any  spherical  triangle  whose  parts 
are  all  less  than  180°  ;  then  the  remainder  or  complement 
of  the  sphere  is  also  a  spherical  triangle  whose  sides  are 
a,  b  and  c,   and  whose  angles  are  360°  —  A,  360°  —  _B, 
and  360°  —  C.     We  shall  distinguish  these  triangles  from 
each  other  by  means  of  accents,  writing  the  letters  within 
the  triangle   to   which    they   respectively   belong,   as    in 
Fig.  14.     The  sides  are  common,  but  when  referred  to  as 
sides  of  A'B'(7,  they  will  be  denoted  by  a',  b'  and  c'  '. 
one  of  the  sides  may  exceed  180°,  as  the  side  a  of  the  triangle  ABC, 
In  this  triangle,  it  is  evident  that  we  must  have  A>180°,  so  long  as  B, 
c  are  each   <  180°.     In  the  triangle  A'B'C'  we  have  ^4/<180°,   while 
,  C">180°. 

FIG.  15.  FIG.  16. 

B' 


Again, 
Fig.  15. 
C,b  and 


If  we  next  suppose  two  of  the  sides  to  exceed  180°,  as  a  and  b,  Fig.  10,  these  sides 
intersecting  in  two  points  whose  distance  is  180°,  the  figure  ceases  to  present  the 
214 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.       215 

triangle  as  an  enclosed  surface,  but  it  will  presently  appear  that  such  triangles  are 
solved  by  the  same  general  methods  that  apply  in  other  cases.  To  form  a  just  con- 
ception of  the  triangle  in  this  case,  we  may  conceive  Fig.  16  to  be  obtained  from 
Fig.  15  by  carrying  the  point  A  along  the  arc  OA  produced  until  it  crosses  the  side  a; 
the  points  A  and  B  may  then  be  joined  either  by  an  arc  less  than  180°,  as  in  Fig.  16, 
or  by  its  supplement  to  360°,  as  in  Fig.  17,  in 
which  last  case  every  side  exceeds  180°.  In 
these  figures,  to  avoid  confusion,  the  point  A 
is  not  placed  in  its  true  position  according  to 
perspective. 

In  each  figure  we  have  two  triangles,  whose 
sides  are  common,  and  whose  angles  are  sup- 
plements to  360°.  It  will  be  easy  to  trace  the 
two  triangles  signified  by  A  BC  and  A'B'C',  by 
remarking  that  the  letters  in  each  case  are  all 
on  the  same  side  of  the  perimeter  of  the  triangle. 

We  may  go  farther,  and  suppose  the  arc  joining  A  and  B  to  be  a  circumference 
-f-  the  arc  AB,  or  any  number  of  circumferences  +  AB;  and  similarly  the  angles  may 
be  supposed  to  be  altogether  unlimited ;  but  since  the  relative  positions  of  any  three 
points  of  the  sphere  must  be  fully  determined  by  arcs  and  angles  less  than  360°, 
nothing  is  gained  by  passing  beyond  this  limit. 

111.  All  the  formuke  of  Chapter  I.  are  applicable  to  the  general  spherical  triangle. 

This  proposition  might  be  considered  as  established  by  the  principle  of  PI.  Trig. 
Art.  49,  but  it  is  also  very  easily  established  by  a  continuation  of  the  process  of  Spher. 
Trig.  Art.  6,  where  the  fundamental  equation  was  shown  to  apply  to  all  triangles  whose 
parts  are  less  than  180°. 

It  was  proved  in  Art.  29,  that  all  the  equations  of  Chap.  I.  may  be  deduced  from  the 
fundamental  one, 

cos  o  =  cos  b  cos  c  -f-  sin  6  sin  c  cos  A  (M) 

We  have  then  only  to  prove  the  generality  of  this  single  equation. 

1st.  Let  all  the  sides  be  <  180,  but  A/  >  180°,  Fig.  14.  The  formula  being  true  for 
the  triangle  ABC,  we  have 

cos  o  =  cos  6  cos  c  +  siQ  b  sin  c  cos  (360°  —  A') 
or  in  the  triangle  A'B'C',  by  PL  Trig.  (76), 

cos  a'  =  cos  V  cos  c'  -f-  sin  bf  sin  c'  cos  A' 

2d.  Let  o>180°,  Fig.  15,  and  produce  a  to  complete  the  great  circle.  The  triangles 
ABO  and  A'B'C'  are  respectively  the  difference  and  sum  of  a  hemisphere  and  the 
triangle  A'ik,  all  of  whose  parts  are  <  180°.  In  the  triangle  A'ik  we  have,  in  terms 
of  the  parts  of  ABC, 

cos  (360°  —  o)  =  cos  6  cos  c  +  sin  6  sin  c  cos  (360°  —  A] 
and  in  terms  of  the  parts  of  A'B'C', 

cos  (360°  —  a')  =  cos  V  cos  c'  +  sin  b'  sin  c,'  cos  A' 

both  of  which  reduce  to  the  form  (M).  But  it  is  here  necessary  to  show  that  the 
formula  may  also  be  applied  to  each  of  the  other  angles:  thus  the  triangle  A'ik 
gives 

cos  b  =  cos  (360°  —  a  )  cos  c  +  sin  (360°  —  a  )  sin  c  cos  ( 180°  —  B) 
cos  b'  =  cos  (360°  —  a')  cos  c/  -f  sin  (360°  —  a')  sin  c/  cos  ( B'  —  180°) 

both  of  which  reduce  to  the  form  (M). 


216  SPHERICAL  TRIGONOMETRY. 

3d.  Let  a  >  180°,  b  >  180°,  Fig.  18 ;  these  arcs  intersect  at  i,  and  the  triangle  A'B'i 

gives 

FIG.  18. 


cos  (a  —  180°)  =  cos  (6  —  180°)  cos  c  +  sin  (6  —  180°)  sin  c  cos  (360°  —  A) 
cos  (a'  —  180°)  =  cos  (bf  —  180°)  cos  c'  +  sin  (V  —  180°)  sin  c7  cos  A' 

which  reduce  to  the  form  (M)  ;  and  in  the  same  way  the  formula  applies  to  the  angle 
B.    We  have  also 

cos  e  =  cos  (a  —  180°)  cos  (b  — 180°)  +  sin  (a  —  180°)  sin  (6  —  180°)  cos  i 

and  since  cos  t  =  cos  C=  cos  (360°  —  C'}  =  cos  C",  this  also  reduces  to  the  form  (M) 
for  both  A  B  C  and  A'  B'  C'. 

4th.  Let  a  >  180°,  b  >  180°,  e  >  180°,  Fig.  19 ;  the  side  c  being  produced  to  com- 
plete the  circle,  the  triangle  i  k  I  gives 

FIG.  19. 


C> 


cos  (a  —  180°)  =  cos  (6  — 180°)  cos  (360°  —  c)  +  sin  (b  —  180°)  sin  (360°  —  c)  cos  I 

and  since  cos  I  =  cos  (180°  —  A)  =  —  cos  A  =  cos  (A'  —  180°)  =  —  cos  A',  this 
reduces  to  the  form  (M)  for  both  ABC  and  A'  Bf  Gf  ;  and  in  the  same  way  the  for- 
mula applies  to  the  angle  B.  We  have  also 

cos  (360°  —  c)  =  cos  (o  — 180°)  cos  (b  —  180°)  +  sin  (a  —  180°)  sin  (b  —  180°)  cos  i 

and  since  cos  i  =  cos  (7=  cos  Cf,  this  reduces  to  the  form  (M)  for  both  ABC  and 
A'  B'  C'. 

The  cases  in  which  the  angles  or  sides  ezceed  360°  are  included  in  the  preceding,  in 
consequence  of  PL  Trig.  Art.  45. 

112.  The  preceding  demonstration,  though  tedious,  has  the  advantage  of  giving  a 
definite  conception  of  the  figures  which  our  formulae  represent.  But  perhaps  the  most 
satisfactory  (as  it  is  the  most  elegant)  method,  is  to  rest  the  demonstration  of  our 
fundamental  equations  themselves  upon  the  principles  of  analytical  geometry,  and,  for 
the  sake  of  those  who  are  acquainted  with  that  subject,  we  add  the  following 
investigation : 

Any  point  of  the  sphere  may  be  referred  by  rectangular  co-ordinates  to  three 
planes  passing  through  the  centre  of  the  sphere  at  right  angles  to  each  other.  Let 
0  be  the  centre  of  the  sphere,  Fig.  20,  and  A  B  C  &  spherical  triangle  upon  its 
surface.  Let  one  of  the  co-ordinate  planes,  as  X  Y,  coincide  with  the  great  circle 
A  B,  and  let  the  axis  of  X  pass  through  B.  HOP  be  drawn  perpendicular 


SOLUTION  OF    THE  GENERAL  SPHERICAL  TRIANGLE.      217 


to  the  plaue  -X"  Y,  and  CP'  and  PP'  to  the  axis  OX,  the  co-ordinates  of  the  point 
G  are 

z  —  0  P',        y  =  PP',        z=CP 


FIG.  20. 


FIG.  21. 


the  values  of  which  (0  C being  taken  =  1)  are 

x  =  cos  a 

y  —  sin  a  cos  B 

z  =  sin  a  sin  B 


s     Ofi' 


If  now  the  axis  of  -X"  be  made  to  pass  through  A,  Fig.  21,  without  changing  the 
position  of  the  plane  X  Y  we  shall  have  for  xf,  y',  z',  the  co-ordinates  of  C  referred 
to  the  new  axes, 

xf  =      cos  b 

yf  =  —  sin  6  cos  A 

z'  =      sin  6  sin  A 

The  axis  of  z  being  unchanged,  the  relations  between  x',  yf,  and  z,  y,  are  expressed 
simply  by  the  formulae  for  the  transformation  of  co-ordinates  in  a  plane  ;  the  in- 
clination of  the  new  axes  to  the  first  is  here  expressed  by  c,  and  the  formulae  of  trans- 
formation are  therefore 


x  =  x'  cos  c  —  y  sin  c 
y  —  xf  sin  c  +  yf  cos  c 
z=z' 

substituting  the  values  of  the  co-ordinates,  we  have  at  once  the  three  following  funda- 
mental equations : 


cos  a  =  cos  c  cos  6  +  sin  c  sin  6  cos  A 
sin  a  cos  B  =  sin  c  cos  b  —  cos  c  sin  5  cos  A 
sin  a  sin  B  =  sin  6  sin  A 


(N) 


which  are  identical  with  (4),  (6),  and  (3). 

113.  Having  established  the  complete  generality  of  our  fundamental  equations,  we 
may  now  employ  for  the  solution  of  the  general  triangle  any  of  those  deduced  from 
them  in  Chap.  I. 

As  a  single  trigonometric  function  is  not  sufficient  to  determine  an  unlimited  angle 
or  arc,  (PI.  Trig.  Art.  53),  it  becomes  necessary  in  most  cases  to  deduce  expressions 
for  both  the  sine  and  cosine  of  the  required  part. 

It  will  be  found  that  all  the  six  cases  of  the  general  triangle  admit  of  two  solutions,  but 
that  they  all  become  determinate,  when,  in  addition  to  the  other  data,  the  sign  of  the  sine  or 
cosine  of  one  of  the  required  parts  is  given.     In  the  practical  applications  in  astronomy, 
it  mostly  happens  that  the  conditions  of  the  problem  supply  this  sign. 
28  T 


218 


SPHERICAL  TRIGONOMETRY. 


114.  CASE  I.  Given  6,  c  and  A.  First  Solution  ;  when  one  of  the  remaining  angles, 
as  B,  and  the  third  side  a  are  required.  The  relations  between  the  given  and  required 
parts  are 


cos  a  =  cos  c  cos  6  +  sin  c  sin  6  cos  A 
sin  a  cos  B  =  sin  c  cos  6  —  cos  c  sin  b  cos  A 
sin  a  sin  B  =  sin  6  sin  A 


(196) 


The  signs  of  the  second  members  will  be  known  from  their  computed  numerical 
values  ;  the  sign  of  cos  a  is  therefore  known.  If  the  sign  of  sin  a  is  also  given,  the 
quadrant  in  which  a  must  be  taken  will  be  known ;  the  second  and  third  equations 
will  determine  the  sign  of  the  sine  and  cosine  of  B,  and  therefore  the  quadrant  in 
which  B  is  to  be  taken. 

In  like  manner,  if  the  sign  of  either  cos  B  or  sin  B  is  given,  that  of  sin  a  becomes 
known,  and  the  problem  is  determinate.  If  no  conditions  are  attached  to  the  required 
parts,  there  must  be  two  solutions. 

The  numerical  solution  will  be  conducted  as  follows :  The  values  of  the  second 
members  (or  simply  their  logarithms)  are  to  be  separately  computed,  and  their  signs 
carefully  noted  ;  then  the  quotient  of  the  3d  by  the  2d  (or  the  difference  of  their 
logs.)  will  give  tnn  B,  and  hence  B,  which  will  be  taken  in  the  quadrant  indicated  by 
the  signs  of  the  sine  and  cosine.  Then  the  3d  divided  by  sin  B,  or  the  2d  by  cos  B, 
will  give  sin  a,  which,  agreeing  with  the  value  from  the  1st  equation,  will  serve  to 
verify  the  correctness  of  the  whole  process. 

This  solution  may  be  adapted  for  logarithms  by  the  methods  employed  in  the  pre- 
ceding chapter. 

1st.  Let  k  and  <j>  be  determined  by  the  equations 


It  sin  $  =  sin  b  cos  A 
k  cos  0  =  cos  b 


k  being  a  positive  number  (PI.  Trig.  Art.  174) ;  then 

cos  a  =  k  cos  (c  —  0) 
sin  a  cos  B  =  k  sin  (c  —  <J>) 
sin  a  sin  B  =  sin  b  sin  A 


(197) 


2d.  Eliminating  k,  and  taking  0<  180°,  (PI.  Trig.  Art.  174), 

cos  A        (<t 
cos  (c  —  0) 


tan  (j>  —  tan  b  cos  A 
cos  6 


sin  a  COB  B  =  ^-°— -  sin  (c  —  0) 

cos  <j> 

sin  a  sin  B  —  sin  b  sin  A 


(198) 


3d.  If  the  quadrant  in  which  &  is  to  be  taken  is  given,  we  may  give  the  preceding  equa- 
tions the  following  form : 


tan  0  =  tan  b  cos  A 
tan  a  cos  B  =  tan  (c  —  0) 

D       sin  0  tan  A 
tan  a  sin  B  =  — 

cos  (c  —  0) 


OK  180°) 


(199) 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.       219 

4th.  If  both  a  and  b  are  less  than  180°,  as  not  unfrequently  happens  iu  the  appli- 
cations of  this  problem,  let 


sin  a 
n  =  — 


(200) 


sin  b  k 

then  m  and  n  are  both  positive  numbers  (k  being  positive)  and  (197)  gives 

TO  sin  ft  =  cos  A 
m  cos  ft  =  cot  6 
n  sin  B  =  sin  ft  tan  A 
n  cos  B  =  sin  (c  —  ft) 
cos  a  =  cot  (c  —  ft)  cos  B 

Check.  We  find 

sin  (c  —  ft) sin  o  cos  B tan  A 

sin  ft  sin  6  cos  A      tan  B 

cos  (c  — ft) cos  a 

cos  ft  cos  6 

besides  which  we  may  employ,  in  connection  with  ( 200),  the  equation  sin  o  sin  B  = 
sin  6  sin  A;  or  in  connection  with  (197)  or  (198)  the  equation  tan  a  cos  B  = 
tan  (c  —  ft).  Or  when  (197)  and  (198)  are  employed,  we  may  find  a  both  by  its  sine 
and  its  cosine. 

115.  The  angle  C  may  be  found  in  the  same  manner  as  B,  interchanging  B  and  C, 
b  and  c,  in  the  preceding  formulae.  But  when  B  and  O  are  both  required,  the  Second 
Solution  to  be  given  presently  is  preferable. 


EXAMPLE. 

Given  A  =  261°  16',  b  =  45°  54',  c  =  138°  32',  and  a  <  180° ;  to  find  a  and  B. 

We  shall  first  employ  (197).  The  first  column  of  the  following  computation,  con- 
taining the  symbols  expressing  the  operations  to  be  performed,  should  be  prepared 
before  opening  the  tables : 

A  261°  16' 
6  45°  54' 
c  138°  32' 

log  sin  A  —  9-9949352 

log  cos  A  —  9-1813744 

log  sin   b  +  9-8562008 

log  cos  6  =  log  k  cos  ft  -f  9-8425548 

log  sin  6  cos  A  =  log  k  sin    <t>  —  9*0375752 

log  tan   ft  —  9-1950204 

log  cos  <t>  +  9-9947336 

log  k  +  9-8478212 

ft  351°   5'42"-6 

*  c  —  <t>  147°  26'17"-4 


*  As  <t>  >  c,  we  take  c  =  138°  32r  +  360°,  so  that  c  —  0  may  be  a  positive  angle  ; 
but  it  would  be  equally  convenient  to  take  c  —  <t>  =  —  212°  33X  42//'6. 


320  SPHERICAL  TRIGONOMETRY. 

log  sin  (c  —  <t>)    +  9-7309514 

log  cos  (c  —  0)    —  9-9257303 

log  k  cos  (c  —  0)  =  log  cos  a  —  9'7735515 

a  126°25'6"-6 

(1)   log  sin  6  sin  A  =  log  sin  a  sin  B   —  9'8511360 

log  k  sin  (c  —  0)  =  log  sin  a  cos  B   -j-  9'5787726 

log  tan  B  —  0-2723634 

B  298°6'26"-8 

log  sin  a   +  9'9056351 

log  sin  B  —  9-9455009 

(1 )    Check,     log  sin  a  sin  B  —  9'8511360 

If  a  were  not  limited,  we  should  have  two  solutions,  the  second  being  a  = 
233°  34'  53" -4,  B  =  118°  6'  26"  -8. 

We  shall  neit  give  the  computation  by  (200),  which  is  applicable  to  this  example, 
since  both  a  and  6  are  less  than  180°. 

A  261°  16' 

b  45°  54' 

c  138°  32' 

log  cos  A  =  log  m  sin  0  -  -   9'1813744 

log  cot  6  =  log  m  cos  <j>  +   9'9863540 

log  tan  ^  -  -   9-1950204 

0351°    5'42"-6 

c  — 0  147°  26'  17"'4 

log  tan  A  +   0-8135608 

log  sin  ^  --    9-1897534 

log  tan  A  sin  0  —  log  n  sin  B  —   0  0033142 

log  sin  (c  —  0)  =  log  n  cos  B  -f    97309514 

log  tan  B  —  0-2723628 

B  298°6'26//-9 

log  cos  -B  +   9-6731379 

log  cot  (c  —  0)  --   0-1947789 

log  cos  B  cot  (c  —  0)  =  log  cot  a  --  9-8679168 

a  126°25'6"-7 

116.  CASE  I.  Given  b,  c  and  A.  Second  Solution ;  when  the  two  angles  B  and  C, 
or  when  all  the  remaining  parts  are  required.  We  have,  by  Gauss's  Equations  (44), 

cos  £  a  sin  £  (B  -)-  C)  =  cos  $  (b  —  c)  cos  $  A 
cos  i  a  cos  $  (1?  -f  (7)  =  cos  J  (6  +  c)  sin  J  ^4 
sin  J  a  sin  ^  (5  —  C)  =  sin  £  (6  —  c)  cos  J  ^4 
sin  J  a  cos  J  (J?  —  C)  =  sin  J  (6  -f  c)  sin  J  ^ 

From  the  first  two  we  deduce  \  (B  -f-  (7)  and  cos  \  a,  and  from  the  second  two 
i  (B  —  C)  and  sin  J  a,  whence  5,  C  and  a.  The  problem  becomes  determinate,  as 
before  ;  that  is,  when  a  is  limited  by  one  of  the  conditions 

a<180°,     or  a  >  180° 
for  then  the  signs  of  both  cos  J  a  and  sin  J  a  will  be  known.* 

*  By  Art.  27,  Gauss's  Equations  may  also  be  taken  with  the  negative  sign  when  the 
triangle  is  unlimited,  as  in  the  group  (45),  but  the  same  final  results  are  obtained 
from  either  (44)  or  (45).  See  note  at  the  end  of  this  chapter,  p.  227. 


(202) 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.       221 

EXAMPLE. 
Same  as  in  Art.  115.     a<  180°. 

A  261°  16' 

b  45°  54' 

c  138°  32' 

$(b  —  c)  46°  19' 

J  (6  +  e)  92°  13' 

\  A  130°  38' 

log  d  =  log  cos  $  (b  —  c)  +  9-8392719 

log  e  =  log  sin  |  (6  —  c)  —  9'8592393 

log  /  =  log  cos  J  (6  -f  e)  —  8-5874694 

log  g  =  log  sin  J  (6  +  c)  +  9-9996749 

log  cos  }A—  9-8137250 

log  sin  I  A    +  9-8801803 

log  d  cos  J  A  —  log  cos  £  a  sin  J  (!»'  +  C)  —  9'6529969 

log  /  sin  \  A  —  log  cos  \  a  cos  \  (B  -f-  C)  —  8'4676497 

log  tan  }  (B  +  C)  +  1-1858472 


log  sin  \  (S  +  (7)  —  9-9990771 
log  cos  £  a    -f  9-6539198 

Ja   63°12'33"-3 

log  e  cos  $  A  —  log  sin  £  a  sin  J  (5  —  C)  +  9'6729643 
log  gsinl  A  —  log  sin  }  a  COB  }  (1?  —  (7)  -f  9'8798552 
log  tan  j  (B  —  0)  +  9-7931091 

l(B  —  C)  31°50'28"-7 

log  sin  $  (B  —  C)  +  97222788 
log  sin  £  a     +  9'9506855 
la    63°12'33"'3 

=  298°   6'26"7 
Ans.      1  C  =  234°  25'  29"'3 
a  =  126°  25'   6"-6 

117.  If  a  only  is  required,  we  may  find  it  by  one  of  the  methods  of  Arts.  75  and 
76  ;  and  if  the  sign  of  sin  a  is  given,  the  solution  is  determinate.     If  the  sign  of  sin 
B  or  of  sin  C  is  given,  we  find  that  of  sin  a  by  inspecting  the  equation 

•       _  sin  A  sin  6  _  sin  A  sin  c 
sin  B  sin  C 

118.  CASE  II.  Given  A,  C  and  b.     First  Solution;  when   the  third   angle  B,  and 
one  of  the  remaining  sides  (as  «)  are  required. 

The  general  relations  between  the  given  and  required  parts  are 

cos  B  =  —  cos  C  cos  A  -\-  sin  C  sin  A  cos  6         ~) 
sin  B  cos  a  =       sin  C  cos  A  +  cos  Csin  A  cos  6          (•        (203) 
sin  B  sin  o  =  sin  A  sin  6          j 

which  are  solved  in  the  same  manner  as  (196).    The  problem  is  determinate  when 
the  sign  of  either  sin  £,  cos  a,  or  sin  a  is  given. 

T2 


222 


SPHERICAL  TRIGONOMETRY. 


Adapting  (203)  for  logarithms,  we  find 
1st. 

*  k  sin  &  =  cos  A 
k  cos  tf  =  sin  .4  cos  b 

cos  B  =  k  sin  (C—&) 
sin  B  cos  a  =  k  cos  ( C  —  &) 
sin  J5  sin  a  —  sin  A  sin  6 


(k  positive) 


2d. 


cot  i?  =  tan  A  cos  6 


cos  B 


sn  £  cos  a  = 


COs  (G  — 


sin  B  sin  a  =  sin  A  sin  6 
3d.    When  the  quadrant  in  which  B  is  to  be  taken  is  given  : 

cot  #  =  tan  A  cos  6 
tan  B  cos  a  =  cot  (C  —  $) 

tanJ3sina=    tan  6  cos  * 
sin  (<7— #) 

4th.  JP&en  A  and  B  are  both  less  than  180°,  let 
k 


9  = 


then  p  and  q  are  positive  numbers,  and  we  have  from  (204), 

p  sin  $  =  cot  J. 
p  cos  i?  =  cos  6 
q  sin  a  =  tan  6  cos  # 
g  cos  a  =  cos  (C  —  i?) 
cot  5  =  tan  ((7—  tf)  cos  o 


Checks.     We  have 


cos  ((7  —  tf)  _  sin  jB  cos  a  _  tan  6 
cos  * 


sin  A  cos  6 
sin  (C  —  #) cos  B 


tan  a 


(204) 


(205) 


(206) 


(207) 


(208) 


*  The  same  factor  k  is  used  here  and  in  (197),  although  the  auxiliaries  0  and  $  are 
different.  To  show  that  k  lias  the  same  value  in  (197)  and  (204),  let  the  squares  of 
the  equations 

k  sin  0  =  sin  6  cos  A  k  cos  0  =  cos  b 

be  added  ;  we  find 

P  (sin*  0  -f  cos5  $)  =  If  —  cos2  6  +  sin2  6  cos2  ^4  =  1—  sin2  6  sin2  A 
and  in  the  same  manner,  from  the  equations 

k  sin  i?  —  cos  A  k  cos  •&  =  sin  -4  cos  6 

we  find 

k*  =  1  —  sin"  6  sin2  A 

and  therefore,  in  both  cases,  k  =  \/  (1  —  sin2  6  sin2  A) 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.       223 

besides  which  we  may  employ,  with  (207),  the  equation  sin  B  sin  a  =  sin  A  sin  6; 
or  with  (204)  and  (205),  the  equation  tan  B  cos  a  =  cot  (G — #).  Also,  when  (204) 
or  (205)  is  employed,  we  may  find  B  both  by  its  sine  and  its  cosine. 

These  formulae  are  computed  in  the  same  manner  as  those  of  preceding  case. 

119.  CASE  II.  Given  A,  O  and  6.  Second  Solution;  when  the  two  sides,  a  and  c, 
or  all  the  remaining  parts,  are  required.  We  employ  Gauss's  Equations  in  the  fol- 
lowing form  : 

sin  J  B  sin  £  (a  +  c)  =  cos  £  ( A  —  C)  sin  £  b 
sin  \  B  cos  J  (a  +  c)  =  cos  }  (A  +  C)  cos  J  b 
cos  *  B  sin  J  (a  —  «)  =  «»  J  (^  —  c')  sin  t  6 
cos  i  5  cos  $  (a  —  c)  =  sin  \  (A  -\-  C)  cos  J  b 

which  are  solved  in  the  same  manner  as  (202). 

EXAMPLE. 

Given  A  =  121°  36'  19"'8,  C  =  42°  15'  13"-7,  6  =  40°  V  10",  and  5  >  180°. 

By  (209). 


b 

40°    0'10"-0 

A 

121°  36'  19"-8 

c 

42°  15'  13"  -7 

\(A-0) 

$(A  +  C) 

39°  40'  33"-0 
81°  55'  46"-7 
20°    0'    5"-0 

log  d  =  log  cos  £  (.4  —  C) 
log  e  —  log  sin  ^  (A  —  C) 
log  /  =  log  cos  $  (A  -{-  C) 

+  9-8863038 
+  9-8051224 
+  9-1473326 
+  9-9956775 

log  sin  $  6 
log  cos  J  6 

+  9-5340806 
+  9-9729820 

log  <Z  sin  J  6  =  log  sin  }  B  sin  J  (a  -f  c) 
log  /  cos  J  6  =  log  sin  J  5  cos  3  (a  +  c) 
log  tan  J  (o  -j-  c) 

+  9-4203844 
+  9-1203146 
4-  0-3000698 

J  (a  4-  c) 

63°  23'  3"-3 

log  sin  J  (a  +  c) 
log  sin  £  5 

4-  9-9513526 
4-  9-4690318 

|1* 

162°  52'  28"-6 

*log  (  —  e  sin  $  6)  =  log  (  —  cos  £  B)  sin  J  (a  —  c) 
*log  (  —  g  cos  }  6)  =  log  (—  cos  j  5)  cos  £  (a  —  c) 
log  tan  J  (a  —  c) 

—  9-3392030 
—  9-9686595 
4-  9-3705435 
193°12'32"-9 

f-       

256°  35'  36"-2 
230°  10'  30"'4 

- 

325°  44'  57"'2 

*The  sign  of  each  of  these  factors  is  changed  because  B  >  180°,  and  cos  J  B  is 
negative. 

f  It  was  necessary  to  increase  J  (a  -f-  c)  by  360°,  to  obtain  e.  The  corresponding 
value  of  6  would  be  616°  35'  36"'2.  See  note  at  the  end  of  this  chapter,  p.  227. 


224 


SPHERICAL  TRIGONOMETRY. 


120.  When  B  only  is  required,  we  may  employ  the  methods  of  Arts.  81  and  82, 
which  are  determinate  when  the  sign  of  sin  B  is  given ;  or  when  that  of  either  sin  a 
or  sin  c  is  given,  since  we  may  then  find  that  of  sin  B  by  inspecting  the  equations 

n       sin  A  sin  b       sin  C  sin  b 

sin  .o  = = . 

sin  a  sin  c 

121.  CASE  III.  Given  a,  6  and  A.     First  Solution;  when  the  three  remaining  parts 
B,  C,  and  c  are  all  required. 

We  find  B  by  the  equation 

(210) 


which  is  determinate  when  the  sign  of  cos  B  is  given. 
Then,  to  find  C,  we  have 

—  cos  C  cos  A  -j-  sin  C  sin  A  cos  b  =  cos  B 
sin  Ccos  A  -j-  cos  (7 sin  .4  cos  6  =  sin  B  cos  a 

which  have  already  been  employed  and  adapted  for  logarithms  in  Art.  118.     If  we 
denote  the  auxiliary  by  $,  and  put  C —  &  =  #',  we  find,  from  (204), 

k  sin  #  =  cos  A  (k  positive) 

k  cos  &  =  sin  .4  cos  6 

A  sin  $'  =  cos  B 

k  cos  #'  =  sin  5  cos  a 


To  find  c,  we  have 


(211) 


cos  c  cos  6 
sin  c  cos  6 


8m  c  sm     cos  -    =  cos  a 
cos  c  sin  b  cos  .4  =  sin  a  cos 


which  have  already  been  employed  and  adapted  for  logarithms  in  Art.  113.     If  we 
denote  the  auxiliary  by  ^,  and  put  c  —  <j>  =  0',  we  find,  from  (197), 

jt  sin  ^  =  sin  6  cos  A  (k  positive) 

k  cos  0  =  cos  6 


Checks.     We  have 


k  sin  0'  =  sin  a  cos 
A  cos  $'  =  cos  a 

cos  A  cos 


(212) 


cos  # x 


sn 
sin 


tan 


tan  a 
tan  6 

cos  b 


tan  A 


(213) 


One  of  which  may  be  used  as  a  check  when  either  C  or  c  has  been  alone  computed.1 
When  both  C  and  c  have  been  found,  the  obvious  check  is 


sin  C      sin  A 


(214) 


*The  following  relations  deserve  a  passing  notice: 
8in  *  «•  *r  =  sin  5  sin  B 


cos  <t>  sin  0' 


=  sin  a  sn 


tan  »  tan  ^  _  g; 
tan  i?  tan  #  ' 

sin  2  <t>  sin  2 


sin1  6 
sin  2  1?  sin  2  <j>'      sin2  a 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.        225 


EXAMPLE. 


Given  a  =  126°  25'  6"'6,  b=  138°  32'  0",  A  =  261°  16'  0",  and  cos  B  negative. 


126°  25'  6"-6 
138°  32'  0"'0 
261°  16'  0"'0 


By  (210), 


By  (211), 


log  sin  a  +  9'9056351 

log  sin  b  +9-8209788 

log  sin  A  —  9-9949352 

log  sin  B  —  9-9102789 

B  234°25'29"-3 

log  cos  6  —9-8746795 

log  sin  A  cos  6  =  log  k  cos  &  +  9'8696147 


--  log  k  sin 
log  tan 


9-1813744 
&  —  9-3117597 
tf  348°24'53"-0 


By  <212), 


log  cos  a  —  9-7735515 

log  sin  B  cos  a  =  log  k  cos  #'  +  9-6838304 

log  cos  B  —  log  k  sin  #'  —  97647520 

log  tan  &'  —  0-0809216 

$'  309°41'33"-7 

#  +  #'  =  C  298°  6'26"-7 

log  sin  6  cos  A  —  log  *  sin  </>  —  9'0023532 

log  cos  b  =  log  k  cos  <t>  —  9'8746795 

log  tan  <t>  +  9-1276737 

<t>  187°38'31"-3 


log  sin  a  cos  B  ==  log  k  sin  $ 
log  cos  a  =  log  It  cos  <t>' 
log  tan  $ 


-  9-6703871 

-  9-7735515 
+  9-8968356 
218°15/28//-6 

45°53/59/A9 


log  sin  C    —  9-9455010 
log  sin  c     -f  9-8562006 


log 


sin  c  i 


Check. 


-  0-0893004 


—0-0893001 


In  this  example,  both  i?  -f  i?x  and  <&  -f-  <j>'  exceed  360°,  and  consequently  we  have  to 
deduct  360°  from  each  of  them.  We  might  have  avoided  this,  however,  by  taking 
$'  =  —  50°  18'  26//-3,  <t/  =  —  141°  44'  31"'4. 

122.  CASE  III.  Given  a,  b  and  A.  Second  Solution;  when  C  and  c  are  required 
without  finding  B. 

We  have  only  to  eliminate  B  from  the  fourth  equation  of  (211)  by  means  of  (210), 
and  then  (omitting  the  third  equation)  determine  #'  by  its  cosine,  observing,  however, 
to  take  it  so  that  sin  #'  shall  have  the  sign  of  cos  B,  which  sign  is  supposed  to  be 
given.     The  formulae  for  finding  C  thus  become 
29 


226 


SPHERICAL  TRIGONOMETRY. 

k  sin  #  =  cos  A  (k  positive) 

k  cos  #  =  sin  A  cos  6 
cos  #'  =  cos  $  cot  a  tan  6 
(#'  <  180°  with  the  sign  of  cos  B) 


(215) 


To  find  c,  we  observe  that  sin  <t>'  has  the  sign  of  sin  a  cos  B,  so  that  we  have  the  fol- 
lowing- formulae : 

k  sin  0  =  sin  b  cos  A         (k  positive) 
k  cos  0  =  cos  6 

(216) 


cos  6 
(<t>'  <  180°  with  the  sign  of  sin  a  cos  B) 


Check.     The  equation  (214). 

123.  CASE  IV.  Given  A,  B  and  6.     First  Solution;  when  the  three  remaining  parts 
a}  e  and  Care  all  required. 
We  find  a  by  the  equation 

•  sin  A  sin  6 


sin  B 


(217) 


which  is  determinate  when  the  sign  of  cos  a  is  given.     The  remainder  of  the  solution 
is  by  (211)  and  (212). 

124.  CASE  IV.  Given  A,  B  and  6.     Second  Solution;  when  c  and  C  are  required, 
without  finding  a. 
We  easily  find,  from  (211), 


k  sin  &  =  cos  A 

k  cos  &  =  sin  A  cos  6 

a, sin  #  cos  JS 

sin  w    —  - 

cos  .4 


positive) 


(cos  $'  and  sin  -B  cos  a  to  have  the  same  sign) 


(218) 


And  from  (212), 


k  sin  0  =  sin  6  cos  A  (k  positive) 

k  cos  $  =  cos  b 

sin  ?/  =  sin  <t>  tan  ^4  cot  5 
(cos  d'  and  cos  a  to  have  the  same  sign ) 

c  =  <(>  +  +' 

Check.    The  equation  (214). 

125.  CASE  V.  Given  a,  6  and  c.     The  formula 

cos  a  —  cos  b  cos  c 
cos  A  —  — 

sin  o  sin  c 


(219) 


(220) 


determines  A  when  the  sign  of  sin  A  is  known.     If  the  sign  of  sin  B  or  of  sin  C  is 
given,  that  of  sin  A  becomes  known  by  the  equation 

sin  A sin  B sin  C 

sin  a        sin  6        sin  c 


SOLUTION  OF  THE  GENERAL  SPHERICAL  TRIANGLE.       227 

The  formulae  (31),  (33),  (34),  may  be  used,  each  of  which  will  become  determinate 
when  the  sign  of  either  sin  A,  sin  B,  or  sin  C  is  known. 
126.  CASE  VI.     Given  A,  B  and  C.     The  formula 

(221) 


sin  B  sin  C 

determines  a  when  the  sign  of  sin  o  is  given.  If  the  sign  of  sin  6  or  of  sin  c  is  given, 
that  of  sin  a  becomes  known  by  the  equation 

sin  a   _  sin  b  _  sin  c 
sin  A       sin  B      sin  O 

The  formulae  (36),  (38),  (39),  may  be  used,  each  of  which  will  be  determinate  when 
the  sign  of  either  sin  a,  sin  b,  or  sin  c  is  known. 

NOTE  UPON  GAUSS'S  EQUATIONS. 

In  the  unlimited  spherical  triangle,  we  may  consider  any  part,  as  a,  to  have  an 
infinite  number  of  values,  viz.  a,  a  -f  360°,  a  +  720°,  etc.,  expressed  generally  by  the 
formula  a  -f-  2  n  TT,  n  being  any  whole  number  or  zero  ;  and  since 

sin  a  =  sin  (o  -f-  2  n  TT)  cos  a  =•  cos  (o  -f-  2  n  K) 

all  those  equations  of  Chap.  I.  that  involve  only  sin  a  and  cos  a  will  not  be  changed  by 
the  substitution  of  a  +  2  n  TT  for  a.  A  similar  substitution  may  be  made  for  each  of 
the  parts,  or  for  all  of  them,  at  the  same  time,  so  that  there  is  an  infinite  series  of  tri- 
angles to  which  these  equations  are  applicable. 

But  the  substitution  of  a  -f  360°  for  a,  in  Gauss's  Equations,  (202),  will  change  the 
sign  of  all  of  them,  since 

sin  J  (o  +  360°)  =  —  sin  J  a  cos  J  (o  +  360°)  =  —  cos  J  a 

while  the  substitution  of  a  +  720°  for  a  will  not  change  their  sign,  since 

sin  J  («  +  720°)  =  sin  }  a  cos  £  (a  -f  720°)  =  cos  J  o 

In  general,  their  sign  is  changed  by  the  substitution  of  o  -f-  (4  n  -f-  2)  ir  for  a,  and  it  is 
not  changed  by  the  substitution  of  a  4-  4n  ir.  The  same  results  follow  like  substitu- 
tions for  each  of  the  parts.  It  follows  that  these  equations  taken  only  with  the  posi- 
tive sign,  do  not  include  all  the  triangles  of  the  infinite  series  above  spoken  of,  and 
that  they  are  complete  only  when  taken  with  the  double  sign,  and  expressed  in  two 
distinct  groups,  as  (44)  and  (45)  of  Art.  27. 

In  practice,  however,  we  may  take  them  with  the  positive  sign  only  ;  for  they  will  then 
give  at  least  one  of  the  triangles  of  the  series,  from  which  .ill  the  others,  (and  particu- 
larly that  whose  parts  are  less  than  360°),  may  be  directly  deduced  by  the  application 
of  360°.* 

This  will  be  illustrated  by  the  example  of  Art.  119,  p.  223  ;  we  there  find 

}  (a  +  c)  =    63°  23/    3"'3 
}  (a  —  c)  =  193°  12'  32"-9 
or  rather,  since  £  (a  -(-  c)  should  be  greater  than  £  (a  —  c), 

$  (a  +  c)  =  423°  23'    3"  3 
£  (a  —  c)  =  193°12'32"-9 

*  Gauss  (Theoria  Molus  Corp.  Ccel.  Art.  54)  recommends  the  use  of  the  positive  sign 
only,  observing  that  any  side  or  angle  may  be  diminished  or  increased  by  360°,  as  the 
case  may  require,  but  confines  himself  to  the  statement  of  this  practical  precept,  with- 
out explaining  the  grounds  upon  which  it  rests. 


228  SPHERICAL  TRIGONOMETRY. 

whence 

a  =  616°  35'  36"-2 

c  =  230°  10'  30/A4 

which  is  the  proper  solution  of  the  equations  taken  with  the  positive  sign.     If  now  we 
deduct  360°  from  a,  and  take,  as  on  p.  223, 

o  =  256°35/36"-6 
c  =  230°  10'  30"-4 

we  have  the  solution  that  would  have  been  obtained  by  taking  the  negative  sign  in  all 
the  equations ;  for  we  now  have 

$  (a  -f  c)  =  243°  23'    3"'3 
J  (a  —  «)=    13°  12/32/A9 

which,  differing  from  the  former  values  by  180°,  must  change  the  sign  of  all  the 
equations. 

I  have  given  some  further  particulars  respecting  unlimited  spherical  triangles,  and 
a  fuller  discussion  of  Gauss's  Equations,  in  an  essay  which  the  reader  will  find  in  the 
Astronomical  Journal,  Vol.  I.,  published  at  Cambridge,  Mass. 


CHAPTER   V. 

AREA  OF  A  SPHERICAL  TRIANGLE. 

127.  Given  the  three  angles  of  a  spherical  triangle,  to  compute  the 
area. 

This  problem  is  solved  in  geometry,  where  it  is  proved  that  the 
surface  of  a  spherical  triangle  is  measured  by  the  excess  of  the  sum  of 
its  three  angles  over  two  right  angles,  by  which  is  meant,  that  the  area 
is  as  many  times  the  area  of  the  tri-rectangular  triangle  as  there  are 
right  angles  in  the  excess  of  the  sum  of  the  angles  over  two  right  angles. 

To  express  this  analytically,  let 

r  =  radius  of  the  sphere 

T—  surface  of  the  tri-rectangular  triangle 

=  ^  surface  of  a  sphere  =  ^nr2 
2S=A  +  B  +  C 
K=  area  of  the  triangle  ABC. 

Also,  let  the  angles  A,  B  and  C  be  expressed  in  the  unit  of  Art.  11, 
that  is,  let  A,  B,  C  denote  the  arcs  which  measure  the  angles  in  a 
circle  whose  radius  is  unity.  The  right  angle  expressed  in  the  same 

unit  is  ->  therefore  the  number  of  right  angles  in  2  8  is 
2 

2S^-  =  — 
2        it 

and  we  have,  according  to  the  above  theorem  of  geometry, 


—  ic) 
x 

or  K=i*(2S—  it)  (222) 

and  if  the  radius  of  the  sphere  is  taken  =  1 

K=2S—  it  (223) 

128.  In  a  plane  triangle  the  sum  of  the  angles  is  equal  to  TT,  and 
in  a  spherical  triangle  the  sum  exceeds  it  by  K  ;  hence  this  quantity, 
Kt  is  commonly  called  the  spherical  excess. 

U  229 


230  SPHERICAL  TRIGONOMETRY. 

129.  Given  the  three  sides,  to  find  the  area. 
By  (223),  we  have 

sin  J  K=  sin  Is  —  ^J  =  —  cos  S 


=  cos  IS  —  -  J  =  sin<S 


cos          = 

tan  $  A'=  —  cotS 


(224) 


in  which  we  have  only  to  substitute  the  values  of  cos  S,  sin  S,  and  cot  S,  given  in 
Art.  34,  to  obtain  the  required  solution.     We  find,  [s  =  J  (a  -f-  6  -{-  c)], 


.     ,  jr \/  [sin  a  sin  (s  —  a)  sin  (s  —  6)  sin  (8  —  c)] 

2  cos  \  a  cos  $  6  cos  \  c 

i  ,rr cos  a  4-  cos  &  4"  cos  c  4~  1 

c08  z  -*>•  —  — : ; TT 1 — 

4  cos  i  a  cos  J  6  cos  $  c 

cos*  ^  a  4-  cos8  j  b  4~  cos*  ^  c  —  1 

2  cos  J  a  cos  £  6  cos  £  c 

The  numerator  of  (225)  being  denoted  by  n,  we  find, 

,  ,  JT-      14-  cos  a  4-  cos  6  4"  c08  c 

COt  *  A  — 

2n 
which  is  known  as  De  Gua's  formula. 


(225) 


(226) 


(227) 


Again,  from  the  formulae  of  Art.  35,  since  1  —  sin  5  =  2  sin  *\  K,  1  -f- sin  S  — 
2  cos  *  J  K,  we  find 


sn          = 


cos     a  cos       cos     c 
=       fcos  }  a  cos  £(«  —  a)  cos  H»  —  6)  cos  }  (»  —  c) 


K  =  v/  fc 
L 


cos  \  a  cos  J  6  cos  £  r,  J 

tan  J  -ST=  i/  Ltan  J  s  tan  i  (s  —  a)  tan  $  (s  —  ^)  tan  i  (s  —  c)3 

the  last  of  which  is  known  as  LhuiUier's  formula. 

130.   Given  two  sides  and  the  included  angle,  (or  a,  6  and  C)  to  find  the  area. 
We  have,  from  (224),  by  (71), 


(228) 


.  i   ,£-  _ 


cos  C 


sin  C 


i    !>•  tan  i  «  tan  i  6  sin  C 

tan  5  A  =  — 

1  -f-  tan  J  (i  tan  j  6  cos  C 


(229) 


AREA  OF  A  SPHERICAL  TRIANGLE.  231 

131.  If  we  admit  more  than  three  parts  of  the  triangle  into  the  expression  of  K,  we 
have,  by  (56)  and  (67), 


i  -fT- cos  $  a  cos  \  b  -f-  sin  ^  a  sin  $  b  cos  (7 

cos  5  A  — — -        —* • 


(230) 


the  quotient  of  which  gives  (229). 

132.  Since  there  are  always  two  triangles  upon  the  surface  of  the  sphere  which  have 
the  same  three  sides,  (Art.  110),  the  angles  not  being  limited  to  values  less  than  180°, 
the  formula  (225),  (226),  (227)  should  give  the  areas  of  both  of  them,  and  their  sum 
should  be  equal  to  the  surface  of  the  sphere  —  4  ir.  In  fact,  by  (225),  sin  £  K  may 
be  either  positive  or  negative,  while  by  (226)  the  cosine  is  fully  determined,  so  that 
these  formulae  give  two  values  of  i  K  whose  sum  is  2  TT,  and  therefore  two  values  of  A", 
whose  sum  is  4  TT. 

It  follows  that  (225)  alone  is  not  sufficiently  determinate  when  the  triangle  is  un- 
limited, since  it  gives  four  solutions.  The  most  convenient  formula  is  therefore  (228), 
for  we  must  always  have  J  K  <  TT,  and  the  double  sign  of  the  radical  gives  the  two 

values  of  J  K,  one  less  and  the  other  greater  than  — 


CHAPTER    VI. 

DIFFERENCES  AND  DIFFERENTIALS  OF  SPHERICAL  TRIANGLES. 

133.  Two  parts  of  a  spherical  triangle  being  constant,  and  a  third 
receiving  an  increment,  it  is  required  to  deduce  the  corresponding 
increments  of  the  remaining  three  parts.      As  in   plane  triangles, 
(PI.  Trig.  Chap.  XII.),  this  will  be  effected  by  a  comparison  of  two 
triangles  having  two  parts  in  common.     The  triangle  formed  from 
the  given  one  by  applying  the  increments  to  the  variable  parts  will 
be  distinguished  as  the  derived  triangle. 

We  shall  first  consider  the  increments  as  finite  differences,  and  give 
them  the  positive  sign,  (PL  Trig.  Art.  187). 

134.  CASE  I.    A  and  c  constant.      The  parts  of  FlG-  *%, 
ABC,  Fig.  22,  being  A,  c,  B,  C,  a,  6,  those  of  the 

derived  triangle  ABO'  are  A,  c,  B  +  J.B,  C+  JC, 
a-i-Ja,  6  +  J6;  and  the  parts  of  the  differential 
triangle  BCC'  are  a,  a  +  Ja,  J6,  180°  —  (7,  <?+  JC 
and  AB.  We  have,  then,  in  BCC',  by  (3), 

sin  J6  _  sin  (a  -f-  A  a)  _         sin  a  ,~~~  . 

sin~J  £  ~        sin  C  ~          sin(C+JC) 

Also,  in  BCC',  by  (40),  we  have 

sin  4-  (180°—  C-f  C-}-  AC]  =  tan  ^  Jfe 

sin  £  (180°  —  C—  C—  JC)  ~~  tan  £  (o  -f  Ja  —  a) 

whence 

tan  ^  Ja  _  cos(C-f^C) 


By  (41)  we  find  in  a  similar  manner, 
tan  ^  J6  _  _  tan(a-f 


By  (42), 

sin  ^  Ja  _  sin  (q  H~ 
~ 


232 


DIFFERENCES  OF  SPHERICAL  TRIANGLES.  233 

By  (43), 

tan  ^  JO  _  __  cos  (a  +  \  Ja) 
tan  \  AB  cos  £  Ja 

By  combining  (232)  and  (233), 

tan  \  Ja  __  _     tan  (a  -f  ^  Ja)  (2361 

tan£  JO  =         tan(O-f  £  JO) 

As  these  formulae  involve  the  increments  in  the  second  members, 
they  are  to  be  computed  by  successive  approximations.  (See  PI. 
Trig.  Art.  201). 

135.  CASE  II.  A  and  a  constant.  The  given 
triangle  being  A  B  C,  Fig.  23,  the  parts  of  the 
derived  triangle  A'  B  C  are  A,  a,  B  +  A  B,  b  -f  J6, 
O+  JO,  c  +  Jc.  Although  the  figure  appears  to 
show  that  the  angle  B  is  diminished,  it  is  still  proper 
to  represent  the  angle  A'  B  C  by  B  +  ABt  to  preserve  uniformity  in 
the  algebraic  signs  of  the  increments  ;  the  essential  signs  being  given 
by  the  equations  of  differences  themselves.  Hence  we  put  the  angle 
ABA'  =  ABC-A'BC=B-(B+  AB)  =  —  AB.  Joining^' 
we  have  in  BA  A'  and  OA  A',  by  (43), 

cos  (c  +  |  Jc)  :  cos  £  Jc  =  -  cot  |  AB  :  tan  |  (A'  A  B  -f  ^4  4'  -B) 
cos  (6  +  J  J6)  :  cos  i  J6  =      cot  |  JO  :  tan  ±(A'AC  +  AA'C} 

but  since  -4  is  constant,  or  B  A  C=  B  A'  C,  we  find  that  the  fourth 
terms  of  these  proportions  are  equal  ;  whence 

tan  |  J.B  _  _  cos  (6  +  £  J6)cos  \  Jc 
tan  £  JO  cos  (c  +  }  Jc)  cos  £  J6 

In  the  polar  triangle  of  A  B  C}  the  constants  are  still  an  angle  and 
its  opposite  side,  and  the  preceding  equation  applied  to  this  polar 
triangle  (by  Art.  8)  gives 


ten  $  J6         _  cos(Jg  +  ^-  J7?)cos^-  JO  ,„     , 

tan^Jc   "          cos(O+|  JO)  cos  £  JJ5 

In  -<4  _K  O  and  ABC  we  have 

sin  a  sin  B  =  sin  -4  sin  6 
sin  a  sin  (B  +  JJ5)  =  sin  ^1  sin  (6  -f-  J6) 

30  u  2 


234  SPHERICAL  TRIGONOMETRY. 

the  difference  and  sum  of  which  give 

sin  a  cos  (£  -f  |  AB)  sin  \  A  B  =  sin  A  cos  (b  -f  ^  J6)  sin  |  J6 
sin  a  sin  (5  +  £  J.Z?)  cos  |  JJ?  =  sin  A  sin  (6  -f  $  Ab)  cos  £  Ab 

from  which,  by  division,  we  find 

tan     J  ten6 


and  in  the  same  manner 

tan|Jc  _.   tan  (c 


, 


tan^JC         tan(C+|JC) 
The  product  of  (237)  and  (239)  gives 

sin  \  Ab   _  _          sin  (6  +  \  Ab]  cos  \  Ac  .    .  . 

tan  ±  JO  ~        cos  (c  +  ^  Jc)  tan  (B  +  $  AB) 

whence  also  * 


(       . 


tan  $  AB  cos  (6+  |  J6)  tan  (O+     AC) 

'  136.  CASE  III.  b  and  c  constant.  The  given 

triangle  being  ABC,  Fig.  24,  the  parts  of  the 
derived  triangle  ABC'  are  6,  c,  a  +  A  a,  B-\-  AB, 
C-f  JC,  A  +  AA.  Joining  CO'  we  have  in  BCC'9 

s  by  (42), 

sin  (a  +  i  4  a)  :  sin  \Aa  =  cot  £  J£  :  tan  \  (B  C  C'  —  B  C'  C) 
But  observing  that  A  C'  =  A  C,  A  C  C'  =  A  C'  C,  we  have 

BCC'  =  ACCr  —  C 
BC'C=AC'C+C+AC 
%(BCC'-BC'C)=    -(C+^JC) 

and  the  above  proportion  gives,  therefore, 

sin  \Aa  =        sin  (a  -f  ±  Aa)  ,       . 

cot(C+£JC) 


*  The  equations  (239),  (240),  (241),  and  (242),  contain  each  two  factors  less  than 
the  corresponding  equations  given  by  Cagnoli. 


DIFFERENCES  OF  SPHERICAL  TRIANGLES.  235 

In  the  same  manner  we  should  find 

sm\Ja_      sin  (a  +  \  Aa) 
tan  $  AC       ~~  c 

The  quotient  of  (243)  and  (244)  gives 


tan^JC       tan(C+£JC) 
In  A  B  C  and  A  B  C',  by  (4),  we  have 

cos  a  =  cos  6  cos  c  -f-  sin  6  sin  c  cos  A 
cos  (a  +  Ja)  =  cos  6  cos  c  -f  sin  b  sin  c  cos  (A 

the  difference  of  which  gives 

sin  ^  A  a  _  sin  b  sin  c  sin  (A  -f" 


sin  ^4A  sin  (a  + 

The  quotient  of  (243)  divided  by  (246)  gives 

sin^  A  A  =    _  sin2  (a  +  £  A  a)  tan  (C+  ^  J  C) 


/'94fi>\ 


tan  ^  J  jB  sin  b  sin  c  sin  (A  + 

and  from  (244)  and  (246),  in  the  same  manner, 

sinJLJJ[^        sin2  (a  -f  £  A  a]  tan  (Jg 
tan^JC  sin  6  sin  c  sin  (  J.  + 

137.  CASE   IV.  B  and  (7  constant.     The  equations  of  the  preced- 
ing case  (243  to  248),  applied  to  the  polar  triangle,  give 

sin(^i-f  |  AA) 

~ 


tan  ^  Ab        cot  (c  -f-  \  Ac) 

sin^  A  A  _  sin(^ 
tan  ^  Ac        cot(b 

tan  1  Ab      tan  (6 


—  -       -  —  — 
tan  ^Jc      tan  (c  +  %  Ac) 


sin  B  sin  Csin  (a  -f-  ^  A  a) 
sin  %  A  a  ~  sin  "(A  + 


236  SPHERICAL  TRIGONOMETRY. 

sin|-  A  a sin2  (A  +  j  A  A)  tan  (c  -f  \  A 

tan  \Ab  sin  B  sin  C  sin  (a  -f  -|  Ad) 


sin  ^  Ja sin2  ( A  -f  1  ^^)  tan  (6  -f  ^ 

tan^Jc  sin  ^  sin  (7 sin  (a  -f- 


FINITE  DIFFERENCES  OF  SPHERICAL  RIGHT  TRIANGLES. 

138.  All  the  preceding  equations  are,  of  course,  applicable  to  right 
triangles,  or  to  quadrantal  triangles,  and  in  some  cases  they  assume 
simpler  forms.     Thus  in  Case  I.,  if  the  variable  O=  90°  (231)  and 
(232)  become 

sin  Ab  =  sin  (a  -f-  A  a)  sin  A  B 
tan  %  Aa  =  —  tan  \  Alt  tan  \  AC 

and  similar  modifications  take  place  in  other  cases. 

139.  When  one  of  the  constants  is  90°,  the  preceding  equations  do 
not  generally  assume  any  simpler  forms,  but  they  may  be  trans- 
formed so  as  to  involve  the  same  variables  in  both  members,  which  is 
generally  desirable  in  their  practical  applications.* 

The  method  that  we  shall  follow  is  so  simple  that  it  will  be  un- 
necessary to  repeat  it  in  every  case.  A  single  example  will  suffice  to 
explain  it. 

Let  C  (=  90°)  and  6  be  the  constants;  to  find  the  relation  of  Ac 
and  AB,  we  have  between  the  two  variables  and  the  constant  6,  the 
equations 

sin  B  =  sin  b  cosec  c 

sin  (B  -\-  AB)  =  sin  b  cosec  (c  +  ^c) 

the  difference  and  sum  of  which,  by  PI.  Trig.  (105),  (106),  (131),  and 
(132),  are 


IT>  .    1    AT>\   •    i    A  n  2  sin  6  cos(c+  \  Ae)  sin  \  Ac 

2  cos  (B  +  A  A  B)  sm  i  AB  =  - 

smcsm(c+  Jo) 

o   •    /  T>  .    1   A  r>\        1   A  T>  2  sin  6  sin  (c  +  i-  Jc)  cos  4-  Jc 

2sm(J5  +  i  JJ5)cosA  AB=      - 

sm  csm(c+  Jc) 


*  Cagnoli  gives  these  equations  reduced  so  as  to  involve  the  same  variables  in  both 
members ;  but  in  almost  every  instance  his  formulae  involve  two  factors  more  than  are 
necessary,  and  are  far  less  simple  and  convenient  than  those  here  given. 


DIFFERENCES  OF  SPHERICAL  TRIANGLES.  237 

and  the  quotient  of  these  is 

tan  4  AB  tan  i-  Ac 

^ £l 

tan  (c  -f  ^  Ac) 


which  gives  the  first  equation  of  the  following  article.     This  process 
always  eliminates  the  constant,  and  is  applicable  in  every  case. 

When  the  equation  to  be  differenced  involves  cosines,  we  employ 
PI.  Trig.  (107)  and  (108);  if  tangents,  (115)  and  (116);  if  cotan- 
gents, (122) ;  if  secants,  (129)  and  (130).  The  results  are  as 
follows : 

140.  CASE  I.    C=  90°  and  6  constant. 

tan^Jc  _         tan(c  +  ^-  Ac)        tan|^  Ac  _     .    cot  (c-)-^  Ac)     /occ\ 
tan%AB~    ~tan(B  +  %AB)       tan±Aa~       cot(a  +  ^Ja)     ^       ' 

sin  Aa  sin  (2  a -|-  A  a)        tan  ^  A  a,  _        tan  (a  -f  j  A  a)     /9t-R\ 

lfa~AA  '         sin(2.A  +  J^)        sin  AB  ~     ~sm(2B-\-AB) 

sin  Ac  sin  (2 c  -j-  Ac) 


141.  CASE  II.    C=  90°  and  c  constant. 

sin  A  A  sin(2A-j-  A  A)       tan^Aa  _        cot  (a-f  j  Aa) 

sinAB  s\n(2B-\-AB)       tan|J6  cot(6-{- 


tan  ^  Aa  _       tan(q+  ^  A  a)       tenjM^.  _        tan  (6  +  \  Ab} 

(>iO9j 


sinJq  __  sm(2a+Aa)        J™Jb_.  sin  (26+  Ab)  ._fi  , 

142.  CASE  III.    C  =  90°  and  A  constant. 

tan^  Ac  __  tan(c+  |  Ac)          sin  Aa  sin  ( 2  a  +  J  a)  ,      . 

tan^Ja  tan(a+|Ja)         tan|J6  tan(6  +  JJi) 

*?B_I^  —  cot(c  +  ^-  Ac)        ian^Ab  _  cot (6  +  ^-  J6)  ,_     . 

Sin  J.B  gJTi^9   7?_L  ,f  Z?\             *«~  1    /<!?""  — i.(T?    \     -\    AT>\  V    "    / 


^HL^_=       sin  (2  c  +  Ac}        tanJ-Ja=      cot^a  ±J-^«)     /2fioN 
sinJ6  sin  (26 -f  J6)        tan^J^B  tan(B-\-^AB) 

143.  If  a  constant  side  is  90°,  the  equations  of  finite  differences 
for  the  triangle  may  be  obtained  by  applying  the  preceding  equations 
to  the  polar  triangle. 


238  SPHERICAL  TRIGONOMETRY. 

DIFFERENTIAL  VARIATIONS  OF  SPHERICAL  OBLIQUE  TRIANGLES. 

144.  To  obtain  the  differential  variations,  we  have  only  to  make 
the  increments  infinitely  small  in  the  equations  of  finite  differences, 
observing  the  principles  of  PI.  Trig.  Art.  192.  Or  we  may  differ- 
entiate. the  equations  of  spherical  triangles  directly,  employing  the 
differentials  of  the  trigonometric  functions  given  in  PI.  Trig.  Art. 
192.  For  example,  A  and  c  being  constant,  to  find  the  relation  of 
d  a  and  d  E,  we  have 

sin  A  sin  c  =  sin  a  sin  (7 
the  differential  of  which  is 

0  =  sin  a  d  sin  C-f  sin  Cd  sin  a 
=  sin  a  cos  Cd  C  -f-  cos  a  sin  Cd  a 

da  _         tan  a 
JC~    ~  tan  C 

and  to  find  the  relation  of  da  and  d  6,  we  have 

cos  a  =  cos  b  cos  c  -f-  sin  6  sin  c  cos  A 
—  sin  ada=  —  sin  6  cos  c  d  b  +  cos  b  sin  c  cos  Adb 

da  __  sin  6  cos  c  —  cos  b  sin  c  cos  A 
db  sin  a 

or  by  (7), 

d  a  ^ 

—  -  =  cos  C 

a  6 

results  which  agree  with  those  found  from  (236)  and  (232),  by  making 
Ja,  Ab  and  JC  infinitely  small.  By  either  method,  then,  the  fol- 
lowing equations  may  be  readily  verified. 


145.  CASE  I.    A  and  c  constant. 

da  _      _  tan  a  db  _          sin  a 

dC~        tan  C  dB  sin  C 


(264) 


da  „  db  tana               ,9fif.x 

-  =  cos  C  —  =  ~r~^,              (Zbb) 

db  d  C  sin  C 

d^=  sina^  dC  _coga               (266) 

dJ5  tan  (7 


DIFFERENTIAL  VARIATIONS  OF  SPHERICAL  TRIANGLES.     239 

146.  CASE  II.     A  and  a  constant. 

f*f\a      r£ 

(267) 
(268) 
(269) 

<JUS   U    UlU   ^ 

147.  CASE  III.  b  and  c  constant. 

da  .  ~        d  a 

-  =  —  sin  a  tan  G*  -  —  —  sin  a  tan  B  (270) 

d.5  d  O 

djB  tan  B  da  -     i    •     r,  /<v7-i\ 

-  =       sin  6  sm  C  (271) 

d  O          tan  C 

eZ  ^4.  sin  A 


dB 

cos  6 

d  b 

cos  B 

dO 

cos  c 

d  c 

cos  C 

d  b 

tan  b 

d  c 

tan  c 

dB 

tan^ 

d  O 

tan  C 

d  b 

sin  b 

d  c 

sin  c 

dC 

cos  c  tan  B 

dB 

cos  b  tan  C 

d  B  sin  £  cos  C      d  C  sin  C  cos  jB 

148.  CASE  IV.  B  and  (7  constant. 
dA 


(272) 


=       sin  A  tan  c  -  =       sin  A  tan  6  (273) 

d  •  6  a  c 

d  6  tan  b  dA  „    . 

-  =      sin  ^  sin  c  (274) 

d  e  tan  c  da 

d  a  _  sin  a  d  a  _  sin  a  /07K\ 

eZ  6  sin  6  cos  c          (/  c  sin  c  cos  b 

DIFFERENTIAL  VARIATIONS  OF  SPHERICAL  RIGHT  TRIANGLES. 

The  preceding  may  also  be  used  for  right  triangles;  but  it  may 
be  desirable  to  have  the  same  variables  in  both  members,  as  in 
the  following  formulae  derived  from  those  of  Arts.  140,  141,  and 
142: 

149.  CASE  I.   C=  90°  and  b  constant. 

d  c  tan   c  d  c  _        cot  c  /o'?£\ 

(  ^j  i  D ) 

d  B  tan  B  da  cot  a 

d^a.  _        sin  2  a  d  a  _        2  tan  a  /07'7\ 

i«i  •  I 
sin  "2  A  dB  sin  2  £ 


d  c  _       sin  2  c  ciL4  _        tan  .4.  /O7fi\ 

d..4  2  cot  J.  d  .B  cot  B 


240  SPHERICAL  TRIGONOMETRY. 

160.  CASE  II.    C=  90°  and  c  constant. 

/>rvf     fi 

(279) 
(280) 
(281) 


dA 

sin  2  A 

d  a 

cot  a 

dB 

sin  2  B 

d  b 

cot  b 

d  a 

tan  a 

d  b 

tan  6 

dA 

tan  A 

dB~ 

tanJ5 

d  a 

sin  2  a 

d  b 

sin  2  b 

dB 

2  cot  £ 

dA 

2  cot  ^4 

151.  CASE 

III.     C=90° 

and  A  constant. 

d  c 

tan  c 

d  a 

sin  2  a 

d  a 

tan  a 

d  b 

2  tan  6 

d  c 

2  cot  c 

d  b 

cot  b 

dB 

sin  2  J5 

dB 

cotB 

d  c 

sin  2  c 

d  a 

cot  a 

db 

sin  2  6 

dB 

tan  ^ 

(282) 
(283) 

(284) 

152.  The  differential  variations  are  often  employed  for  approxi- 
mate results,  instead  of  the  equations  of  finite  differences,  when  the 
increments  are  very  small.  The  remarks  of  PI.  Trig.  Art.  203, 
apply  here  also,  but  it  is  not  necessary  to  introduce  the  radius  in 
seconds,  since  all  the  parts  of  a  spherical  triangle  are  expressed  in 
the  same  unit. 

DIFFERENTIAL  VARIATIONS  OF  SPHERICAL  TRIANGLES  WHEN  ALL  THE  PARTS 

ARE  VARIABLE. 
153.  Let  the  equation 

cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A 
be  differentiated,  all  the  parts  being  variable ;  we  find 

sin  a  d  a  =  (sin  6  cos  c  —  cos  6  sin  c  cos  A)  db 
-f-  (sin  c  cos  b  —  cos  c  sin  6  cos  A )  d  e 
-\-  sin  b  sin  c  sin  Ad  A 

Dividing  by  sin  a,  this  becomes,  by  (7)  and  (3), 

d  a  =  cos  C  d  b  +  cos  B  d  c  -{-  sin  6  sin  C  d  A  (285 

and  in  the  same  manner  from  the  2d  and  3d  equations  of  (4)  we  find 

db  =  cos  A  d  c  +  cos  Cd  a  +  sin  c  sin  A  dB  (286) 

dc  =  cos  B  d  a  -f-  cos  A  d  b  -(-  sin  a  sin  B  d  C  (287) 

From  these  three  equations,  any  three  of  the  six  differentials  da,  db,  dc,  dA, 
dB,  dC,  being  given,  the  other  three  may  be  determined  by  the  usual  processes  of 
elimination. 

If  any  one  of  the  parts  be  supposed  constant,  its  differential  will  become  zero,  and 
these  equations  will  assume  simpler  forms.  If  two  of  the  parts  be  supposed  constant, 
we  can  easily  deduce  all  the  equations  of  Arts.  145,  146,  147  and  148. 


CHAPTER    VII. 

APPEOXIMATE  SOLUTION  OF  SPHERICAL  TRIANGLES  IN  CERTAIN 

CASES. 

154.  WHEN  some  of  the  parts  of  the  triangle  are  small,  or  nearly  90°,  or  nearly 
180°,  approximate  solutions  may  be  employed  with  advantage.     These  are  generally 
found  by  means  of  series. 

155.  In  a  spherical  right  triangle  (the  right  angle  being  C),  given  A  and  c,  to  find  b. 
We  have 

tan  b  =  cos  A  tan  e  (288) 

which  is  of  the  form  in  PI.  Trig.  (493),  and  may  therefore  be  developed  by  (495)  and 
(496)  by  putting  x  =  b,  y  =  c,  p  =  cos  A,  whence 

„„!  i  j 
=  —  tan  »  A. 


p  +  1  1  +  cos  A 

and  (495)  and  (496)  become,  [taking  n  =  0  in  (495),  and  n  —  1  in  (496)], 

b—        e  —  tan1  £  A  sin  2  c  +  J  tan*  J  A  sin  4  e  —  etc.  (289) 

b  =  IT  —  c  +  cot1  \  A  sin  2  c  —  J  cot*  J  A  sin  4  c  -f  etc.  (290) 

If  A  is  small,  cos  A  is  nearly  equal  to  unity,  and  6  exceeds  c  by  a  small  quantity, 
which  is  approximately  found  by  one  or  more  terms  of  the  series  (289). 

If  A  is  nearly  180°,  or  cos  A  nearly  =  —  1,  6  exceeds  K  —  c  by  a  small  quantity, 
which  is  found  by  (290). 

For  examples  of  the  mode  of  computation,  see  PI.  Trig.  Art.  255. 

156.  Although  these  solutions  are  termed  approximate,  it  must  not  be  inferred  that 
they  are  less  accurate  in  practice  than  the  direct  solution  of  (288)  by  the  tables  ;  for  the 
logarithmic  tables  are  themselves  only  approximate,   and  the  neglect  of  the  higher 
powers  in  such  series  as  (289)  and  (290)  may  involve  a  less  theoretical  error  than  the 
similar  neglect  of  the  higher  powers  in  the  series  by  which  the  tables  are  computed. 
In  the  examples  of  PI.  Trig.  Art.  255,  the  thousandths  of  a  second  were  found  with 
accuracy,  which  could  not  have  been  effected  by  a  direct  solution  with  less  than  eight 
decimal  places  in  the  logarithms. 

These  considerations  lead  to  the  frequent  employment  of  approximate  solutions  in 
astronomy. 

157.  If  A  and  b  are  given,  to  find  c,  we  have 

tan  c  =  sec  A  tan  6 
which  is  reduced  to  PL  Trig.  (493),  by  putting  x  =  c,  y  =  b,  p  =  sec  A, 


sec  A  -\-  1       1  +  cos  A 
and  the  series  will  be 

c=         6  +  tan2  J^  sin  26  +  J  tan*  J  .4  sin  4  6  +  etc.  (291) 

c  =  IT—  b  —  cot*  \  A  sin  2  6  —  £  cot*  £  A  sin  4  6  —  etc.  (292) 

31  V  on 


242 


SPHERICAL  TRIGONOMETRY. 


158.  Similar  solutions  apply  to  the  equations  of  right  triangles, 

tan  a  =  sin  b  tan  A 
cot  B  =  cos  c  tan  A 
the  last  being  solved  under  the  form 

tan  (90°  —  B)  =  cos  e  tan  A 

We  may  also  compute,  in  the  same  manner,  the  auxiliaries  0  and  #  in  (122)  and 
(134),  so  frequently  employed  in  the  solutions  of  oblique  triangles. 

159.  In  a  right  spherical  triangle,  given  c  and  A,  to  faid  a,  when  A  is  nearly  90°. 
We  have 

from  which  we  deduce 


sin  a  =  sin  A  sin  c 


(293) 


tan  i  (c  —  a)  =  tan*  (45°  —  }  A)  tan  J  (e  +  a)  (294) 

From  this  we  may  find  c  —  a,  which  is  supposed  very  small,  by  successive  approxima- 
tions. For  a  first  approximation,  let  o  =  c  in  the  second  member,  and  find  thence  the 
value  of  c  —  a  and  of  a;  for  a  second  approximation  substitute  in  the  second  member 
the  value  of  a  just  found ;  and  so  on  until  two  successive  values  agree  as  nearly  as 
may  be  desired. 


EXAMPLE. 

Given  A  =  89°,  c  =  87° ;  find  a. 

Here  45°  —  J  A  =  0°  30',  and  for  the  first  approximation  £  (c  +  o)  =  87° 

log  tan  J  (c  -f-  a)      1-28060 

log  tan2  (45°  — M)      5-88172 

ar  co  log  sin  \"     5'31443 

\(c  —  a)  =  299"74  \ogl(c  —  a)     2'47675 

a  =  87°  —  9'  59"'48  =  86°  50/  0"'52 


2o  APPROX. 

3D  APPROX. 

4xH  APPROX. 

!(•+«) 

log  tan  \  (c  -(-  a) 
,      tan'  (45°-  M) 

86°  55'   0" 
1-26868 

86°55/    8" 
1-26899 

1.  1  op  i  c 

86°  55'   8/A17 
1-26900 

1.  1  Q£  i  e 

sin  1" 

lyoio 

log  i  (c  —  o) 
|(e-«) 

c  —  a 
a 

2-46483 
291/'-63 
9'  43"-26 
86°  50'  16"-74 

2-46514 
291//-83 
9''43"-66 
86°  50'  16'A34 

2-46515 
291"-84 
9'  43/A68 
86°  50X  16'A32 

The  direct  solution  of  (293)  gives  a  —  86°  50'  W,  but  cannot  give  the  fractions 
of  a  second  without  tables  of  more  than  seven  figure  logs.  We  have  given  this  pro- 
blem, however,  not  so  much  on  account  of  its  particular  utility,  as  for  the  purpose 
of  introducing  the  method  of  approximation  to  which  it  leads,  and  which  is  often 
employed. 

The  process  here  explained  may  obviously  be  applied  to  any  equation  of  the  form 


sin  x  =  m  sin  y 


when  m  is  nearly  equal  to  unity. 


APPROXIMATE  SOLUTION  OF  SPHERICAL  TRIANGLES.      243 

160.  In  a  spherical  oblique  triangle,  given  two  sides  and  the  included  angle,  to  find  the 
other  angles  and  side  by  series. 

If  a,  6  and  C  are  the  data,  to  find  c,  we  have 

cos  c  —  cos  a  cos  6  -f-  sin  a  sin  6  cos  C 
Substituting  half  arcs, 

sin*  i  c  =  sin2  $  a  cos*  J  b  +  cos*  i  a  sin3  J  6 

—  2  sin  £  a  cos  J  6  cos  £  a  sin  £  b  cos  C 

which  is  of  the  form  PI.  Trig.  (507),  and  may  be  developed  by  (508)  by  substituting 
sin  $  c  for  c,  sin  $  a  cos  £  6  for  a,  and  cos  £  a  sin  £  6  for  6  ;  so  that  (508)  becomes 

,,  ["tan  i  a        ri  •    /tan  \  a\2  cos  2  (7    ,    „  „"!     /oa*\ 

logsin  $c  =  logcosjasm  £6  —  M\         —cos  (7  -f  I—  -  +  etc.       (295) 

|_tan  £  o  Uan  £  o/         Z  J 

To  find  4  and  5,  we  have, 

tan  lA  +  B) 


i  /  t        n\       sin  *  (u  —  o)       ,  i  >-y 
tan  J  (-A  —  -B)  =  —  -^  —  -^  cot  i  (7 
sin  i  (a  +  6) 


Comparing  these  equations  with  PL  Trig.  (493),  and  developing  by  (495),  n  =  0, 

etc.  (296) 


tan  ^  a  Uan  J  a 

If  we  develop  by  (496),  we  find 


cot  J  a  Vcot  J  a/          2 

^.  +etc.  (297) 


S_«\  sin  (7-  (52Li^)  «n_2+  etc  (298) 

tan  ^  &/  vtan 


n  (7-     5L«        H_     _  etc.         (299) 
tan  ^  67  \tan  ^  o/         2 

from  which  a  selection  will  be  made  in  any  particular  case,  according  to  the  con- 
vergency  of  the  series.  The  terms  of  the  series  are  in  arc,  and  must  be  reduced  to 
seconds,  by  dividing  by  sin  V. 

This  solution  may  be  applied  to  the  case  where  two  angles  and  the  included  side  are 
the  data,  by  means  of  the  polar  triangle. 

161.   To  express  the  area  of  a  spherical  triangle  in  series. 

Comparing  (229)  with  PI.  Trig.  (500),  and  developing  by  (502),  we  find 

J  K=  tan  J  a  tan  £  6  sin  C—  J  tan*  J  a  tan*  \  b  sin  2  C+  etc.  (300) 


244  SPHERICAL  TRIGONOMETRY. 

162.  LEGENDRE'S  THEOREM.  If  the  sides  of  a  spherical  triangle  are  very  small  com- 
pared with  the,  radius  of  the  sphere,  and  a  plane  triangle  be  formed  whose  sides  are  equal 
to  those  of  the  spherical  triangle,  then  each  angle  of  the  plane  triangle  is  equal  to  the  corre- 
sponding angle  of  the  spherical  triangle  minus  one-third  of  the  spherical  excess. 

Let  a,  6  and  c  be  the  sides  of  the  spherical  triangle  expressed  in  arc,  the  radius  of 
the  sphere  being  unity;  and  let  A',  Bf  and  G'  be  the  angles  of  the  plane  triangle 
whose  sides  are  a,  b  and  c.  Then  we  have,  in  the  spherical  triangle, 

_    i  _  cos  a  —  cos  6  cos  c 

COS  -o.  •  —  --- 

sin  6  sin  c 

Substitute  in  the  second   member  of  this,  the  values  of  cos  a,   etc.,  in   series,  by 
PL  Trig.  (405)  and  (406),  neglecting  only  powers  above  the  fourth,  viz. 

cos  a  =  1  —  i  a3  +  zV  a* 

cos6  =  l  —  J62  +  ^6*  8in&  =  6  —  $&» 

cos  c  =  1  —  J  c1  +  A  c*  sin  c  =  c  —  £  c* 

we  find 

cos  A  =  *  (**  +  «*-<*')  +  A  (a4-*4  --ct-6ye') 
&  c[l-i(6'  +  <•»)] 

Multiplying  the  numerator  and  denominator  by  1  -f  i  (&1  +  c*),  and  neglecting  terms 
of  a  higher  order  than  the  fourth,  as  before,  we  have 

,  A—  b'  +  c*  —  a'   ,   q*  +  6*  +  c*  —  2a'6»  —  2o*c*  —  26V 

COS  ^.l  -  -  •  —  —  ~T—     ••  '  •  ------      ..  •  •        .  - 

2  b  c  24  6  c 

which,  by  PI.  Trig.  (225)  and  (239),  becomes 

cos  A  =  cos  A'  —  \  b  c  sin2  A' 

Let  A  =  A'  -f  x,  then  since  x  is  small,  we  may  put  cos  z  =  1,  so  that,  by 
PI.  Trig.  (38), 

cos  A  =  cos  A/  —  x  sin  A' 
whence 

x  =  ^  b  c  sin  A/ 

But  J  6  c  sin  A'  =  area  of  the  plane  triangle  =  very  nearly  area  of  the  spherical 
triangle  =  K,  whence 

x=\K  A'  =  A  —  \K 

The  same  reasoning  applies  to  each  of  the  other  angles,  so  that 


which  proves  the  theorem/ 

163.  This  theorem  is  applied  in  geodetical  surveying,  and  is  found  to  be  sufficiently 
accurate  for  triangles  whose  sides  are  considerably  greater  than  1°.  It  is  to  be  remem- 
bered that  the  sides  are  to  be  expressed  in  arc;  and  if  they  are  given  in  fe«t  (for 
example),  they  must  be  reduced  to  arc  by  dividing  by  the  radius  in  feet,  or,  which  is 
equivalent,  the  area  must  be  divided  by  the  square  of  this  radius.  If  then  r  =  radius 
of  the  earth  in  units  of  any  kind,  a,  b  and  c  the  sides  of  the  triangle  in  units  of  the 
same  kind,  and  k  the  area  of  the  plane  triangle,  we  shall  have  K  in  seconds,  by  the 
equation 

jr—       k 

r1  sin  1" 

EXAMPLE. 

In  a  triangle  upon  the  earth's  surface,  given  b  =  183496'2  feet,  c  =  156122'l  feetj 
and  A  ~  48°  V  32"'35;  to  find  the  remaining  parts. 


APPROXIMATE  SOLUTION  OF  SPHERICAL  TRIANGLES.      245 

We  have  k  =  J  6  c  sin  A,  and  the  mean  value  of  r  =  20888780  feet.     Hence 

log  6    5-26363 

log  c     5-19346 

log  sin  A    9-87159 

ar  co  log  2  r1  sin  1"    Q'37356 

K=5"  -04  logJT    0-70224 

It  is  evident  that  great  accuracy  in  the  value  of  r  and  of  the  other  data  is  not 
required  in  computing  K.  We  now  have  j  K  —  1"'68,  A'  =  48°  4'  30"'67,  and  by 
solving  the  plane  triangle  with  the  data  A',  b  and  c,  we  find 

a  =  140580-0  feet  B'  =  76°  12'  22"'19  C'  =  55°  43'  7"13 

Adding  }  K  to  each  of  these  angles,  the  angles  of  the  spherical  triangle  are 
B  =  76°  12'  23"-87  C  =  55°  43'  8"-81 

For  further  details  respecting  geodetical  triangles,  and  for  the  methods  of  solving 
spheroidal  triangles,  special  works  upon  geodesy  must  be  consulted,  such  as  Legendre's 
Analyse  des  Triangles  traces  sur  la  surface  d'une  spheroide  ;  Puissant'  s  Traite  de 
Geodesie  ;  Puissant's  Nouvel  essai  de  trigonometric  spheroidique  ;  Fischer's  Lehrbuch  der 
hoheren  Geoddsie;  various  papers  by  Gauss,  Bessel,  etc. 

164.  To  solve  a  spherical  triangle  when  two  of  ite  sides  are  nearly  90°. 

If  a  and  b  are  nearly  90°,  c  and  0  are  nearly  equal,  and  it  will  be  expedient  to  com- 
pute the  small  quantity  C  —  c  by  an  approximate  method.  We  have,  by  (25), 

sin3  £  c  =  sin'  }(«  +  *)  sin1  }  C  +  sin2  J  (a  —  b)  cos2  }  C 
and  by  PL  Trig. 

sin2  }  0=  [sin2  }  (a  +  6)  +  cos1  J  («  +  6)]  sin2  }  0 
the  difference  of  which  equations  is 

sin  }  (C+c)sin  }  (0  —  c)  =  cos1  }  (o  +  b)  sin2  }  (7—  sin2  J  (0  —  6)  cos1  J  C 
Let 

a'  =  90°  —  a  6'  =  90°  —  b 

a'  and  6'  beiug  very  small  :  also,  since  O  and  c  are  nearly  equal,  put 


then  the  above  equation  becomes 

sin  C  sin  J  (C—  c)  =  sin1  }  (a'  +  6')  sin2  }  (7—  sin*  }  (a'  —  &')  cos2  }  (7 

Dividing  by  sin  (7=2  sin  J  C  cos  £  (7,  and  substituting  the  arcs  J  (O—e), 
J  (a'  +  6'),  j  (a'  —  6'),  for  their  sines,  we  find 

C-  c  =  sin  1"  [  (^^)t  tan  J  C  -  (  a/  ~  6/  )'  cot  J  (?]  (301) 

which  is  the  required  approximate  formula  for  the  case  when  a',  6'  and  C  are  given 
to  find  c. 

If  a',  6'  and  c  are  given,  to  find  (7,  we  may  exchange  C  for  c  in  the  second  member, 
whence 


tan  Je-cotic  (302) 


v  2 


CHAPTER    VIII. 

MISCELLANEOUS  PROBLEMS  OF  SPHERICAL  TRIGONOMETRY. 

165.  In  a  given  spherical  triangle,  to  find  the  perpendicular  from  one  of  the  angles  upon 
the  opposite  side. 

Fl°-  ^  Denoting  the  perpendicular  upon  the  side  c  (Fig.  25) 

C  by  p,  we  have 

sin  p  =  sin  b  sin  A  (303) 

If  the  three  sides  or  the  three  angles  are  given,  we 
find  by  (48),  or  (51),  and  (303), 

A<^T  ^^f  sin  p  =  -7-^-  =  — — -  (304) 

sin  c       sin  C 

in  which  n  and  N  are  given  by  (47)  and  (50). 

If  we  admit  more  than  three  parts  of  the  triangle  into  the  expression  of  p,  we 
have,  by  (55),  (56),  and  (303), 


8n     -    _ 


sn  s  = 
cos  £  C  sin  5  c 


(305) 


166.  To  _/Jnd  <Ae  radius  of  the  circle  described  about  a  given  spherical  triangle. 


FIG.  26. 


The  radius  here  understood  is  the  arc  0  A  =  0  B  =  O  C, 
Fig.  26,  drawn  from  the  pole  of  the  small  circle  A  B  C  to 
either  of  the  angles.  Let 

OAB  =  OB  A  =x 
then         C=  OCA  +  OCB=OA  0+  OBC 

=  A  —  x  +  B  —  x 


putting  S  =  $  (A  -f  B  +6Y  ). 

The  triangle   A  OB  being  isosceles,   the    perpendicular 
0  P  bisecte  the  side  c,  therefore  if  0  A  —  E,  we  have 


or,  by  (66), 


ten  R  =          _  =  __  _ 

cos  x        cos  (  o  —  C  ) 


2  sin  |  a  sin  ^  6  sin  ^  c 


(306) 


By  applying  the  principles  of  Art.  37,  this  will  give  the  corresponding  formulae  of 
PI.  Trig.  (285). 

167.  From  (65)  and  (66)  we  find 

cos  (S  —  C)  —  —  cos  S  cot  ^  a  cot  J  6 
by  which  (306)  is  reduced  to 

tan  £  =  tani«tan}6tanjc 

—  cos  S 
248 


MISCELLANEOUS  PROBLEMS. 
Also,  by  the  last  equation  of  (56),  (306)  becomes 

tan  R  = r~S       A    •     n 

cos  £  a  cos  $  o  sin  C 

168.  Substituting  in  (306)  for  tan  £  c  by  (39), 
tan  R  = 


—  cos  S 


or,  by  (50), 


tan  R 


,cos  (S  —  A)  cos  (S  —  B)  cos  (S—G)l 


—  cos  S 

N 


169.  Let  the  sides  of  the  triangle  ABC,  Fig.  27,  be  pro- 
duced to  meet  in  A',  B',  and  (7  ;  and  denote  the  radii  of 
the  circles  circumscribed  about  A '  B  C,  B'  A  C,  (7  A  B  by 
Rf,  R",  Rf"  respectively.  Then  if  2  S/  denote  the  sum  of 
the  angles  of  A'  B  C,  (A,  B  and  C  being  the  angles  of 
ABC), 

2  S'  =  2  *•  —  B—  C+A 
Sr~A'  =  ir  —  l(A+-B+C)  =  *  —  8 
so  that  (306)  applied  to  A'  B  C  gives 


z?/  _ 


tan  fr  a 


<xx(S'  — 


—  costf 


and  in  like  manner 


-ry,,         tan  \  b 
tan  R"  =  — 

—  cos  o 


Substituting  for  tan  £  a,  etc.,  by  (39),  or  for  cos  S  by  (69), 

p/  __  cos  (S  —  A) 2  sin  j  a  cos  ^  b  cos  ^  c 

R// cos  (S —  B) 2  cos  £  o  sin  $  6  cos  $  c 

JV  n 

?/// cos  (S  —  C) 2  cos  \  a  cos  $  6  sin  \  c 

N  n 


tan 


247 


(309) 


(310) 


FIG.  27. 


(311) 


(312) 


170.  Combining  (310)  with  (312),  we  find  the  relation 

cot  R  cot  R'  cot  R"  cot  -R"7  =  N* 
If  this  be  multiplied  successively  by  the  squares  of  (310)  and  (312),  we  obtain 

tan  R  cot  R/  cot  R"  cot  R"'  —  cos7  S 
cot  B  tan  R'  cot  .R"  cot  B//x  =  cos2  (S—A) 
cot  £  cot  R'  tan  .R"  cot  JR'"  =  cos*  (5  —  J5) 
cot  J2  cot  J2r  cot  E"  tan  JBW  =  cos1  (S  —  C} 


(313) 


(314) 


248  SPHERICAL  TRIGONOMETRY. 

171.  Again,  from  (310)  and  (312)  we  find 

-  tan  R  4  tan  R'  +  tan  R"  +  tan  R"' 

_  cos  S  4  cos  (S  —  A)  4-  cos  (S  —  B)  -f  cos  (S  —  C) 

N 

_  2  cos  }  A  cos  KB  4  C)  4  2  cos  £  ^4  cos  £  (#  —  C) 

N 
whence 

-  tan  R  4  tan  £'  +  tan  R"  +  tan  R"'  =  4  cos  M  cos  $  Jg  cos  }  (7  ^315^ 

We  shall  find  in  a  similar  manner 


tan  R  —  tan  R'  +  tan  JJ"  +  tan  R'"  =  4  cos  M  sin  }  B  sin  j  G 
tan  E  -f  tan  R'  —  tan  jR"  +  tan  R'"  =  4  sin  *  ^  cos  *  ^  sin  *  <? 
tan  ^  +  tan  B'  +  tan  R"  —  tan  JZ'"  =  *«««*}  4  sin  }  J  cos  }  G 


(316) 


It  is  also  easily  shown  that 

tan4  R  +  tan2  .ft'  +  tan1  R"  -f  tan2  R"'  =  2  +  2  cos  A  cos  ^  CO8  G          (317) 

j^1 

172.   To  find  the  radius  of  the  circle  inscribed  in  a  given  spherical  triangle, 

In  Fig.  28,  0  being  the  pole  of  the  required  circle,  draw 
OP7,  OP"  and  OP'"  to  the  points  of  contact,  and  join 
OA,  OB.     We  have  0  P"  =  0  P'"  and  the  triangles 
,    A  O  P"  and  A  0  P"'  right-angled  at  P"  and  P"'  ;  hence 


sn 


B  sin  ^4  0         sin  ^4  0 

therefore  O  A  P"  =  0  J[  Pr//,  (for  we  cannot  have 
0  A  Pff  —  ir  —  0  A  P"f\  and  the  pole  of  the  inscribed  circle  is  consequently  found 
by  the  same  construction  as  in  piano,  namely,  by  bisecting  the  angles  of  the  triangle. 

If  then  we  put  s  =  J  (a  -)-  6  +  c)>   and  *"  =  radius  of  the  inscribed  circle,  we 
have 

A  P'"  +  B  Pf  4-  GPf  =  A  P"'  4  a  =  s,        ^4  P/x/  =  a  —  a 
and  the  right  triangle  A  0  P"'  gives 

tan  r  =  sin  (s  —  a)  tan  J  A  (318) 

corresponding  with  the  formula  of  PI.  Trig.  (288). 
Substituting,  in  (318),  the  value  of  tan  £  A, 

tan  r  _     //sin  (s  —  a)  sin  (s  —  6)  sin  (s  —  c)  \ 
\  V  sin  s  / 

or  tan  r  =  -£-  (319) 

sin  s 

Substituting,  in  (318),  the  value  of  sin  (s  —  a)  given  by  (58), 

tan  r  =  -  -^-  (320) 

2  cos  £  A  cos  f  Jj  cos  £  C 

Also,  by  (51),  we  have  N  ~  £  sin  B  sin  (7  sin  a,  which  reduces  (320)  to 

tan  r  =  Bin  j  JB  Bin  }  C  sin  a  (321 } 

cos  i  ^4 


MISCELLANEOUS  PROBLEMS. 


249 


173.    Let  the  radii  of  the  circles  inscribed  in  the  three  triangles  A'BC,  B'AC, 
C'AB  of  Fig.  27,  be  r',  r"  and  r'ff.     Then  if  s/  denote  the  half  sura  of  the  sides  of 

A'BC,  we  have 

2  s'  =  2  »r  —  6  —  c-fa 

j/  —  a  =  TT  —  J(a  +  6  +  c 
so  that  (318)  applied  to  the  three  triangles,  gives 

tan  r'    =  sin  s  tan  £  A 


tan  7-//   =  sin  s  tan  J  B 
tan  r'ff  =  sin  s  tan  J  C 

Substituting  in  these  the  values  of  tan  £  A,  etc.,  or  of  sin  s, 
n  N 


tan  rf    = 
tanr"  = 


sin  (s  —  a)        2  cos  %  A  sin  £  _B  sin  \  G 

n          = N 

sin  (s  —  b)        2  sin  £  A  cos  i  J5  sin  $  CY 


,  ,.///  _ 

Itlll  T          - 


Also,  by  (321), 


~~        ;  .    -  rt     •       i      j      •       i     «  i    /"* 

sm  (s  —  c)       2  sm  j  .4  sin  £  £  cos  J  C 


,         cos  ^  B  cos  i  C    . 
tan  rr    = — ^ — r1 —  sin  a 


tan,.//  = 


tan  r///  = 


cos  i  ^ 

cos  M  cos 


cos  ^  C 
174.  The  product  of  (319)  and  (323)  gives 


tan  r  tan  r'  tan  r/x  tan  r/x/=  —  -  =  n 

1 


/x/=  —  -  =   1 
n1 


\vlience,  as  in  Art.  170, 


cot  7-  tan  r'  tan  r"  tan  r//x  =  sin1  s 
tan  r  cot  r'  tan  r"  tan  r'"  =  sin*  (s  —  a) 
tan  r  tan  r'  cot  r"  tan  r//x  =  sin*  (s  —  b) 
tan  r  tan  rx  tan  ?-//  cot  r"f  =  sin*  (s  —  c) 

175.  We  find  from  (319)  and  (323),  as  in  Art.  171, 

...       ,    //   i       ,    ///       4  sin  J  a  sin  \  6  sin  ^  c 
—  cot  r  +  cot  r  +  cot  r"  +  cot  r"  —  — 


,    i   ,  ,,  L    /// 

cot  r  —  cot  r'  +  cot  r"  -f-  cot  rx//  = 

,          ,    //   ,       L    /// 
cot  r  -f  cot  r'  —  cot  r"  +  cot  r"f  = 


4  sin  i  a  cos  i  6  cos  i  c 


4  cos  J  o  sin  J  6  cos  ^  c 


.     ..          ,    ,,,      4  cos  \  a  cos  i  6  sin  J  c 
cot  r  +  cot  r'  +  cot  T"  —  cot  r"  =  — 


2  —  2  cos  a  cos  6  cos  c 


(322) 


1-    (323) 


(324) 


(325) 


(326) 


(327) 


250 


SPHERICAL  TRIGONOMETRY. 


176.  From  (309)  and  (321),  we  find 


tan 


=  4  sin  £  A  sin  £  B  sin  J  (7  cos  £  a  cos  J  6  cos     e 


From  (307)  and  the  first  of  (327), 

—  cot  r  +  cot  r'  +  cot  r"  +  cot  r'"  =  2  tan  R 
From  (315)  and  (320), 

—  tan  R  +  tan  .R'  +  tan  H"  +  tan  .R"'  =  2  cot  r 


(328) 


(329) 


(330) 


and   other  similar  relations   are  found   by  comparing  (312)    with  (327),  and   (316) 
with  (323). 

177.  The  following  relations  are  also  worth  remarking. 

If  p  is  the  perpendicular  from  C  upon  c, 


cot  r  sin      == 


sin  }  (7 


178.  The  pole  of  the  circle  inscribed  in  a  spherical  triangle  is  also  the  pole  of  the  circle 
circumscribed  about  the  polar  triangle;  and  the  radii  of  these  circles  are  complements  of  each 
other. 

The  arcs  bisecting  the  angles  of  a  given  triangle  will  evidently  bisect  the  sides  of  the 
polar  triangle,  and  will  be  perpendicular  to  those  sides  respectively ;  the  common 
intersection  of  these  arcs  is  therefore  at  once  the  pole  of  the  circle  inscribed  in  the  first 
and  circumscribed  about  the  second. 

Again,  if  we  join  the  angular  points  of  the  polar  triangle  with  this  common  pole, 
the  arcs  thus  drawn,  being  produced  to  meet  the  sides  of  the  first  triangle,  are  perpen- 
dicular to  those  sides,  and  therefore  pass  through  the  points  of  contact  of  the  inscribed 
circle.  Each  of  these  arcs  =  90°,  and  is  at  the  same  time  the  sum  of  the  two  radii  of 
the  circles  in  question. 

This  latter  property  is  also  obvious  from  the  analytical  expressions  of  the  two  radii. 
By  means  of  it,  we  might  have  deduced  all  the  formula  for  the  inscribed  from  those 
for  the  circumscribed  circle,  or  vice  versa. 

179.  To  find  the  arc  joining  the  poles  of  the  circles  inscribed  in,  and  circumscribed  about 
a  given  spherical  triangle.* 

3-  29.  Let    0  be  the   pole  of  the   circumscribed   circle, 

Fig.  29,  and   0/  that  of  the  inscribed  circle.      Put 
00' =  D;  then 

cos  D  =  cos  A  0  cos  A  0'  +  sin  A  0  sin  A  0'  cos  0 A  O' 
By  Art.  166,  we  have  OAB  —  S—  C,  whence 

OA  0'  =  S  —  C  —  J  A  =  %(B  —  C) 
We  have  also 


COB  AO'  =  cos  O'P  cos  AP  =  cos  r  cos  (s  —  a) 


sin  .40'  = 


sin  0/P 


sin  OMP      sin  %  A 


*Hymer's  Spherical  Trigonometry. 


MISCELLANEOUS  PROBLEMS.  251 

Therefore, 

— cos  -^ —  =  cot  r  cos  (s  —  a)  +  tan  R  cc 

cos  R  sin  r  sin  j  A 

Substituting  by  (319),  (307),  and  (44), 

cos  D _  sin  S  cos  (s  —  a)  -f-  2  sin  \  b  sin  ^  c  sin  ^  (b  -f  c) 

cos  .R  sin  r  n 

sin  a  -f-  sin  b  -\-  sin  c 

~2» 

whence,  by  (53), 

/     co»  D     Y      i 1  +  sin  o  sin  6  -f-  sin  a  sin  c  -f  sin  b  sin  c  —  cos  a  cos  6  cos  c 

Vcos  R  sin  r/  2  n2 

by  PI.  Trig.  (179),      _  /sin  s  +  2  sin  j  a_sin_t_6_sinj_c\» 
\  n  / 

=  (cot  r  -f  tan  J2)2 
cos"  D  =  cos2  (.R  —  r)  +  cos2  R  sin2  r 
sin1  D  =  sin2  (JZ  —  r)  —  cos2  R  sin2  r  (332) 

If  the  inscribed  circle  is  inscribed  in  A'BC,  Fig.  27,  and  its  radius  —  r',  we  have, 
by  a  similar  process, 

sin2  D'  =  sin2  (R  +  r'}  —  cos2  R  sin2  r'  (333) 

180.  To  find  the  equilateral  spherical  triangle  inscribed  in  a  given  circle. 

If  R  —  radius  of  the  given  circle,  and  ^1  =  one  of  the  angles  of  the  equilateral 
triangle,  we  have,  by  (310),  and  PI.  Trig.  Art.  76, 

.,  r>       —  cos  |  A      3  cos  i  A  —  4  cos8 1 .4. 

tan1  R  = e    -  = *- r-r—, \ — 

cos*  j  A  cos3  ^  A 

whence  cos  \A  =  J  ( 3  ,  _  \  (334) 

\  V  4  +  tan2  .R  / 

181.  To  find  the  equilateral  spherical  triangle  circumscribed  about  a  given  circle. 

If  r  =  radius  of  the  given  circle,  and  a  =  one  of  the  sides  of  the  triangle,  we  find 

BinJa  =  J(— r3-— )  (335) 

\  \4  +  cot2  ?•  / 


182.  Given  the  base  and  area  of  a  spherical  triangle,  to  find  the  locus  of  the  vertex. 

FlG-  s0-  Let  a  =  the  given  base,  and  K  =  area  of  ABC,  Fig.  30. 

Produce  AB  and  AC  to  meet  in  ^l7.  Let  O  be  the  pole 
of  the  circle  described  about  A'BC.  The  radius  of  this 
circle  is  given  by  the  first  equation  of  (311),  which,  by 
(224)  becomes 

tanJB'  =  -^i-|;  (336) 

sin  j  K 

The  second  member  of  this  equation,  being  constant  for 
all  the  triangles  of  the  same  base  a,  and  the  same  area  A", 
shows  that  R'  is  also  constant,  and  consequently,  that  the 


252  SPHERICAL  TRIGONOMETRY. 

point  A  is  always  found  upon  the  circumference  of  the  same  small  circle  A'BC.     But 
A  and  A/  being  the  extremities  of  the  same  diameter  of  the  sphere,  A  is  also  found 
upon  a  small  circle,  equal  and  parallel  to  the  circle  A'BC. 
The  perpendicular  distance  (p')  of  0  from  the  base  BC,  is  found  by  the  equation 

,      cos  R' 
cos  p'  =  — 

cos  £  a 

and  the  pole  of  the  locus  of  A  is  in   the  same  perpendicular,  at  a  distance  from 
BO  —K  —  pf  =  p,  whence 

-™*^.  (337) 


The  equations  (336)  and  (337)  determine  the  radius  and  position  of  the  pole  of  the 
required  locus,  which  may  therefore  be  constructed. 

This  elegant  proposition  is  due  to  Lexell. 

183.  To  find  the  angle  between  the  chords  of  two  sides  of  a  spherical  triangle. 

FIG.  31. 


In  Fig.  31,  0  being  the  center  of  the  circumscribed  circle,  the  angle  between  the 
chords  of  the  sides  AC  and  BO  is  half  the  spherical  angle  A  OB.  If,  then, 

Gl  —  angle  between  the  chords  of  a  and  6 
we  have 

cos  Ci  =  cos  AOP  =  sin  OA P  cos  AP 

or,  by  Art.  166,  cos  Gl  =  sin  (S—  C)  cos  J  c  (338) 

By  (72)  this  becomes 

cos  Ci  =  sin  £  a  sin  J  b  +  cos  £  a  cos  J  b  cos  C  (339) 

184.  The  preceding  problem  is  employed  for  geodetical  triangles,  in  which  Ci  differs 
very  little  from  C,  in  which  case  it  is  expedient  to  compute  the  small  difference 
C  —  (7,  =  x.  We  easily  reduce  (339)  to  the  following  : 

cos  GI  =  cos   J  (a  —  6)  cos2  \  C —  cos  J  (a  +  6)  sin2  £  C 

=  cos2  }  C  —  2  sin2  \(a—b)  cos2  }  C  —  sin2  J  C+  2  sin2  J  (a  -f-  6)sinli  (7 
Subtracting  cos  C  =  cos2  $  (7  —  sin2  £  (7,  we  have, 

sin  H (7+  Ci)  sin  }  ((7—  d)  =  sin2  J  (a  +  6)  sin2  }  (7—  sin2  i  (a  —  6)  cos2  *  (7 
or  approximately,  taking 

sin  J  (C1 4-  Ci)  =  sin  C  =  2  sin  J  C  cos  J  C 
and  sin  J  ((7  -  Ci)  =  J  x  sin  I" 

t  being  expressed  in  seconds, 

x  =      ^       sin'  \  (a  +  6)  tan  }  (7  -       !      sin2  J  (<*  —  *)  cot  |  C  (340) 

sin  1"  sin  l//r 


MISCELLANEOUS  PROBLEMS. 


253 


185.  If  a  great  circle  (DE,  Fig.  32)  bisect  the  base  of  a  spherical  triangle  at  right  angles, 
any  great  circle  (FG),    perpendicular  to  it,  divides  tlie  sides  (AC,  BC)  into  segments  whose 
sines  are  proportional;  that  is, 

sin  FA  ;  sin  FC=  sin  GB :  sin  GC  (341) 

Let  P  be  the  pole  of  ED,  (DP  =90°),  and 
PGF  any  great  circle  drawn  through  P,  and 
therefore  perpendicular  to  DE.  Then,  since 

PB  +  PA  =  2  PD  =  180° 

we  have,  by  (3), 

sin  F  sin  FA  =  sin  P  sin  PA  =  sin  Psin  PB 

=  sin  G  sin  GB 
sin  F  sin  FC  =  sin  G  sin  GC 

whence,  by  division,  the  theorem  (341).     The  arc  FG  is  analogous  to  the  parallel  to 
the  base  in  plane  triangles. 

186.  If  two  arcs  of  great  circles,  (AB,  CD,  Fig.  33),  terminated  by  any  circle,  intersect, 
the  products  of  the  tangents  of  the  semi-segments  are  equal  to  one  another;  that  is, 

tan  $  AE  tan  £  EB  —  tan  £  CE  tan  J  ED  (342) 

FIG.  33. 

Let  P  be  the  pole  of  the  circle  DACB.  Join  PE 
and  draw  the  perpendiculars  PF,  PG,  bisecting  the  arcs 
AB  and  CD.  Then  we  have 

co8FE_  cos  PE  _  cos  PE  _  cos  GE 
cosFB      cosPB     cosPD      cosGD 

cosFE—cosFB  _  cosGE  —  cosGD 
coaFE-}-  cos  FB  ~  cos  GE  +  cos  GD 

which,  by  PI.  Trig.  (110),  gives  (342). 

187.  If  three  arcs  be  drarcn  from  the  angles  of  a  spherical  triangle  through  the  same  point, 
to  meet  the  opposite  sides,  the  products  of  the  sines  of  the  alternate  segments  of  the  sides  will 
be  equal. 

Thus,  in  Fig.  34,  we  shall  have 

sin  AB'  sin  CA'  sin  BC'  —  sin  CB'  sin  BA'  sin  AC'  (343) 

For  we  easily  find 

sin  AP       sin  APB' 


sin  CB' 

sin  CP 

sin  CPB' 

sin  CA' 

sin    CP 

sin  CPA' 

sin  BA' 

sin    BP 

sin  BPA' 

sin  BC' 

sin   BP 

sin  BPC' 

sin^lC"      sin 


sin  APC' 


Multiplying  these  equations  together,  the  product  of  the  second  members  is  unity, 
whence  (343). 

The  same  property  is  easily  extended  to  the  segments  of  the  angles. 

188.  It  follows,  that  when  three  arcs  are  drawn  from  the  three  angles,  so  as  to 
satisfy  the  condition  (343),  they  must  intersect  in  the  same  point.  This  occurs  in  the 
same  cases  as  in  plane  triangles,  that  is,  when  the  angles  are  bisected  ;  when  the  sides 
are  bisected;  when  the  three  arcs  are  drawn  from  the  angles  to  the  points  of  contact 

W 


254 


SPHERICAL  TRIGONOMETRY. 


of  the  inscribed  circle ;  and  when  the  three  arcs  are  the  three  perpendiculars  upon 
the  sides. 

The  first  three  of  these  cases  are  obvious.  To  prove  the  last,  if  A',  B'  and  C', 
Fig.  34,  are  right  angles,  we  have 

cos  AB'  ^  cos  CA'    cosBC' cos  AB     cos  CA     cos  BC 1 

cos  GE'  '  cos  EA'  '  cos^C"  ~~  cos  CE  '  cos  BA  '  cos  AC~ 

whence        cos  AB'  cos  CA'  cos  BC'  =  cos  GE'  cos  EA'  cos  AC' 
and  in  the  same  manner  we  find 

tan  AB'  tan  CA'  tan  BCf  =  tan  CB'  tan  BA'  tan  AC' 

The  product  of  these  two  equations  gives  the  condition  (343),  and  therefore  the  per- 
pendiculars intersect  in  the  same  point. 

189.  To  find  the  arc  drawn  from  any  angle  of  a  spherical  triangle  to  a  given  point  in  the 
opposite  side. 

In  the  triangle  PA  A",  Fig.  35,  let  PA'  be  drawn;  we  have 


(344) 


coe  P  A'  sin  A  A"  =  cos  P  A'  sin  (AA'  +  A'A") 

=  cos  P  A'  cos  Af  A"  sin  A  A'  -f  cos  P  A'  cos  AA'  sin  A'A/f 
But  in  the  triangles  P  A  A',  PA'  A"  we  have,  by  (4), 

cos  PA'  cos  A  A'  =  cos  PA     —  sin  PA'  sin  A  A'  cos  PA'  A 
cos  PA'  cos  A'  A"  =  cos  PA"  -f  sin  PA'  sin  A'  A"  cos  PA'  A 
which  snbstituted  above  give 

cos  PA'  sin  A  A"  =  cos  PA  sin  A'  A"  +  cos  PA"  sin  A  A' 
which  determines  P  A',  the  sides  PA  and  PA"  and  the  segments  of  the  side  A  A" 
being  given. 

190.  Let  three  arcs  PA,  P  A',  PA",  Fig.  35,  passing  through  the  same  point  P, 
be  intersected  by  two  others  A  A"  and  B  B"  whose  intersection  is  Q;  we  have  several 
symmetrical  relations  among  the  parts  of  the  figure  which  find  their  application  in 
astronomy. 

Let  the  points  A,  A',  A"  be  given  in  position  by  their  distances  from  Q,  and  put 

A    Q  =  a  A  B    =(3  P   B    =  y 

A'  Q  =  a'  A'  B'  =p'  P   E'  =  y' 

A"Q  =  a"  A"B"=P"  P  B"  =  r" 

By  PI.  Trig.  (171),  we  have 

sin  o  sin  (a'  —  a")  -f-  sin  a'  sin  (a"  —  a)  -f-  sin  a"  sin  (a  —  a')  =  0 
and  in  Fig.  35, 


sin  Q 


sn  y 


MISCELLANEOUS  PROBLEMS. 


whence 

and  similarly 


sin  ft     sm  y"  sin  E" 

sm  a  =  —      •       -*- -— 

sin  )  sm  Q 


.      .     sin  P'    sin  y"  sin  B" 
sm  a.'—  -r^-.  ---  .  ~A  - 
sin  Y  sin  Q 

sin/?"    sin  j"  sin  E"  _ 

in  a"=  —  -       ---  :  —     ~ 


smy 


which,  substituted  above,  give 


sin  /3_:_  ,_,     _„,  j.  sin^  gin  (a  -,_o)  4.  ^££«n  (« 

CTIM  -v/  em  v'' 


(345) 


sin  y  sin  y'  sm  y' 

Again,  if  we  express  (344)  in  the  notation  of  this  article,  it  becomes 
cos  (3+  y)  sin  (a'-a")  +cos  (p'+  y')  sin  (a"-a)+cos  (/J"+  y")  sin  (a-a')=0     (346) 
which,  added  to  (345),  gives 

»-«>)=0  (347) 


tan  y  tan  yx 

191.  If  P  is  the  pole  of  A  Q,  we  have 


tan  yx/ 


and  (345)  and  (347)  both  give 

tan  0  sin  (a'—  a")  +  tan  8'  sin  (a"—  o)  +  tan  p"  sin  (a  —  a')  =  0       (348) 

192.  To  yinii  <Ae  inclination  of  two  adjacent  faces  of  a  regular  polyhedron,  and  ttte  radii 
of  the  inscribed  and  circumscribed  spheres. 

Let  C  and  E,  Fig.  36,  be  the  centres  of  two  adjacent  faces  whose  common  edge  ia 
A  B;  0  the  centre  of  the  inscribed  and  circumscribed  spheres.  Draw  0  D  bisecting 
A  B  at  right  angles;  draw  CD,  ED,  which  will  also  evidently  be  perpendicular  to 
A  B;  and  put 

/  =  inclination  of  the  faces  =  CD  E 

R  =  radius  of  the  circumscribed  sphere  =  0  A  =  OB 

r  •=  radius  of  the  inscribed  sphere          =  0  C=  O  E 

a  =  one  of  the  edges  =  A  B 

m  =  number  of  faces  that  form  a  solid  angle 

n  =  number  of  sides  of  a  face 

Suppose  a  sphere  to  be  described  about  the  centre  0  with 
any  radius,  and  cad  the  triangle  formed  upon  its  surface 
by  the  planes  COD,  CO  A,  A  0  D;  this  triangle  is  right- 
angled  at  d  and  gives 

cos  cad 


cos  c  d  =  ' 


sin  a  cd 


But  cos  c  d  —  cos  COD,  and 


COD  =  90°— C  D  0  =  -  —  *  / 
2 

cad—-}  angle  of  the  planes  0  A  C  and  O  A  E 

_  j_    2  7r_jrr 
2m         m 


SPHERICAL  TKIGONOMETKY. 


sin  — 
n 


Then  from  the  triangles  0  CD,  A  O D,  etc.,  we  find 

r  =  —  tan  i  1  cot  — 
2  n 


(349) 


(350) 


r  =  R  cos  a  c  —  R  cot  a  c  d  cot  c  a  d  =  R  cot  —  cot  — 

n         m 

R  =  ±  tan  J  /tan  - 
2  m 


(351) 


193.   To  _/md  the  surface  and  volume  of  a  regular  polyhedron. 

Let  /=  number  of  faces  of  the  polyhedron  ;  S  —  the  surface,  and  V  =  the  volume ; 
then  the  area  of  each  face  (the  notation  of  the  preceding  article  being  continued)  is 
equal  to 


A  B  X  CD  X  n  =  a'  •  ^  cot  - 
4          n 


whence 


and  since  V=  S  X  J  r, 


S=a*.!^cot^ 
4          n 


. 
24 


(352) 


(353) 


FIG.  37. 


194.  To  find  the  surface  and  volume  of  a  parallelopiped,  given  the  edges  and  their  inclina- 
tions to  each  other. 

Let  0  P,  Fig.  37,  be  a  parallelopiped, 
whose  edges  0  A  —  a,  0  B  =  6,  0  C  =  c, 
and  their  inclinations  B  0  C=  a,  A  OC  =  ft, 
A  0  B  =  7,  are  given. 

The  area  of  any  face,  as  B  C,  is  found  by 
the  formula  6  c  sin  a,  and  therefore  for  the 
whole  surface,  we  have 

S—  2  (6csin  a  -f  acsin/?  -f-  a  6  sin  y)      (354) 
To  find  the  volume,  let   CD  be  the  alti- 
tude, then 

F=base  ABX  CD  =  ab&\ny>X  CD 
Suppose  a  sphere  to  be  described  about   O,  whose  intersections  with  the  planes 
BOO,  AOC,  AOB  and  DOC  are  B'  C'  =  a,  A'  C'  =  /?,  4'J?'  =  7,  and  C'£'. 
The  triangle  A'  C'  D'  is  right-angled  at  /X,  whence 

CD  =  c  sin  C'D'  =  c  sin  /?  sin 
or  by  (46),  if  <r  =  J  (a  + /?  +  7), 


whence 


c 

CD  =  —  —  |/  [sin  CT  sin  (<r  —  o)  sin  (a  —  /?)  sin  (<r  —  y)] 
sin  7 

V=  2  a  6  c  y  [sin  <r  sin  (ff  —  a)  sin  (a  —  /?)  sin  (a  —  y)] 
THE  END. 


(355) 


/EK.  -... 


spherical 


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